小鸟游六花

vanishment this world

发表于 更新于

一、常用模板

1、Kruskal

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// https://www.luogu.com.cn/problem/P3366
using namespace std;
constexpr int maxn = 1000010;
int F[maxn];
int Find(int x){
    return x == F[x] ? x : F[x] = Find(F[x]);
}
struct E{
    int a, b, w;
    bool operator <(const E& e) const{
        return w < e.w;
    }
} edges[maxn];
int main(){
    int n, m;
    scanf("%d%d",&n, &m);
    for(int i = 1; i <= n; ++i) F[i] = i;
    for(int i = 1; i <= m; ++i){
        scanf("%d%d%d", &edges[i].a, &edges[i].b, &edges[i].w);
    }
    sort(edges + 1, edges + 1 + m);
    long long int ans = 0;
    int cnt = 0;
    for(int i = 1; i <= m; ++i){
        int fa = Find(edges[i].a);
        int fb = Find(edges[i].b);
        if(fa ^ fb){
            ans += edges[i].w;
            F[fb] = fa;
            ++cnt;
            if(cnt == n - 1) break;
        }
    }
    if(cnt == n - 1) printf("%lld\n", ans);
    else printf("orz\n");
    return 0;
}

2、最小生成树的唯一性

对于 Kruskal 算法，只要计算为当前权值的边可以放几条，实际放了几条，如果这两个值不一样，那么就说明这几条边与之前的边产生了一个环（这个环中至少有两条当前权值的边，否则根据并查集，这条边是不能放的），即最小生成树不唯一。

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// POJ-1679
using namespace std;
const int maxn = 1000010;
int F[maxn];
int Find(int x){
    return x == F[x] ? x : F[x] = Find(F[x]);
}
struct E{
    int a, b, w;
    bool operator <(const E& e) const{
        return w < e.w;
    }
} edges[maxn];
int solve(){
    int n, m;
    scanf("%d%d",&n, &m);
    for(int i = 1; i <= n; ++i) F[i] = i;
    for(int i = 1; i <= m; ++i){
        scanf("%d%d%d", &edges[i].a, &edges[i].b, &edges[i].w);
    }
    sort(edges + 1, edges + 1 + m);
    long long int ans = 0;
    int cnt = 0, s1 = 0, s2 = 0, last = 0;
    bool flag = true;
    for(int i = 1; i <= m; ++i){
        if(i > last){
            if(s1 != s2){
                flag = false;
                break;
            }
            s2 = s1 = 0;
            for(int j = i; j <= m; ++j){
                if(edges[i].w == edges[j].w)
                    last = j, s1 += Find(edges[j].a) != Find(edges[j].b);
                else break;
            }
        }

        int fa = Find(edges[i].a);
        int fb = Find(edges[i].b);
        if(fa ^ fb){
            ans += edges[i].w;
            F[fb] = fa;
            ++cnt;
            ++s2;
        }
    }
    if(flag && s1 == s2) printf("%lld\n", ans);
    else printf("Not Unique!\n");
    return 0;
}

int main(){
    int T;
    scanf("%d", &T);
    while(T--) solve();
    return 0;
}

3、次小生成树

非严格次小生成树
  • 求出无向图的最小生成树 $T$，设其权值和为 $M$
  • 遍历每条未被选中的边 $e = (u,v,w)$，找到 $T$ 中 $u$ 到 $v$ 路径上边权最大的一条边 $e’ = (s,t,w’)$，则在 $T$ 中以 $e$ 替换 $e’$，可得一棵权值和为 $M’ = M + w - w’$ 的生成树 $T’$.
  • 对所有替换得到的答案 $M’$ 取最小值即可
严格次小生成树

考虑刚才的非严格次小生成树求解过程，为什么求得的解是非严格的？

因为最小生成树保证生成树中 $u$ 到 $v$ 路径上的边权最大值一定 不大于 其他从 $u$ 到 $v$ 路径的边权最大值。换言之，当我们用于替换的边的权值与原生成树中被替换边的权值相等时，得到的次小生成树是非严格的。
解决的办法很自然：我们维护到 $2^i$ 级祖先路径上的最大边权的同时维护 严格次大边权，当用于替换的边的权值与原生成树中路径最大边权相等时，我们用严格次大值来替换即可。

这个过程可以用倍增求解，复杂度 $O(m \log m)$。

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// 严格次小生成树
// https://www.luogu.com.cn/problem/P4180
using namespace std;
constexpr int maxn = 300100;
bool vis[maxn];
int F[maxn];

int Find(int x){
    return x ^ F[x] ? F[x] = Find(F[x]) : x;
}
struct E{
    int a, b, w;
    bool operator <(const E& e) const{
        return w < e.w;
    }
} roads[maxn];

struct Edge {
    int next, to, w;
} edges[maxn * 2];
int heads[maxn], pos = 0;
void addEdge(int a, int b, int w) {
    edges[++pos].next = heads[a];
    edges[pos].to = b;
    edges[pos].w = w;
    heads[a] = pos;
}

template<int N>
struct T120B {
    int arr[N];
    int cnt = 0;
    T120B() {
        for (int i = 0; i < N; ++i) arr[i] = INT_MIN;
    }
    void add(int x) {
        int i = 0, j = N - 2;
        while (i < N && x <= arr[i]) ++i;
        if (i == N) return;
        while (j >= i) arr[j + 1] = arr[j], --j;
        arr[i] = x;
        if (cnt < N) ++cnt;
    }
};
int stp[maxn][20];
// max value
int stm[maxn][20];
// second-largest value
int sts[maxn][20];
int depth[maxn];

void dfs(int u, int fa){
    depth[u] = fa == -1 ? 0 : depth[fa] + 1;
    stp[u][0] = fa;
    for(int i = 1; ~stp[u][i-1] && i < 20; ++i){
        stp[u][i] = stp[stp[u][i-1]][i-1];
        T120B<4> max4;
        max4.add(sts[u][i-1]);
        max4.add(sts[stp[u][i-1]][i-1]);
        max4.add(stm[u][i-1]);
        max4.add(stm[stp[u][i-1]][i-1]);

        stm[u][i] = max4.arr[0];
        int sv = INT_MIN;
        for(int k = 1; k < 4; ++k){
            if(max4.arr[0] != max4.arr[k]){
                sv = max4.arr[k];
                break;
            }
        }
        sts[u][i] = sv;
    }
    for(int i = heads[u]; i; i = edges[i].next){
        int v = edges[i].to;
        if(fa == v) continue;
        stm[v][0] = edges[i].w;
        sts[v][0] = INT_MIN;
        dfs(v, u);
    }
}

int lca(int u, int v){
    if(depth[u] < depth[v]) return lca(v, u);
    for(int i = 19; ~i; --i){
        if(~stp[u][i] && depth[stp[u][i]] >= depth[v]) u = stp[u][i];
    }
    if(u == v) return u;
    for(int i = 19; ~i; --i){
        if(stp[u][i] != stp[v][i]) u = stp[u][i], v = stp[v][i];
    }
    return stp[u][0];
}
// 求u-v路径上不等于val的最大边
int query(int u, int v, int val){
    int res = INT_MIN;
    for(int i = 19; ~i; --i){
        if(~stp[u][i] && depth[stp[u][i]] >= depth[v]) {
            if(stm[u][i] == val)
                res = max(res, sts[u][i]);
            else
                res = max(res, stm[u][i]);
            u = stp[u][i];
        }
    }
    return res;
}

int main(){
    int n, m;
    //memset(vis, 0, sizeof(vis));
    //memset(heads, -1, sizeof(heads));
    pos = 0;
    scanf("%d%d",&n, &m);
    for(int i = 1; i <= n; ++i) F[i] = i;
    for(int i = 1; i <= m; ++i){
        scanf("%d%d%d", &roads[i].a, &roads[i].b, &roads[i].w);
    }
    sort(roads + 1, roads + 1 + m);
    long long int mst = 0, ans = 0;
    int cnt = 0;
    for(int i = 1; i <= m; ++i){
        int fa = Find(roads[i].a);
        int fb = Find(roads[i].b);
        if(fa ^ fb){
            //printf("select (%d %d %d)\n", roads[i].a, roads[i].b, roads[i].w);
            addEdge(roads[i].a, roads[i].b, roads[i].w);
            addEdge(roads[i].b, roads[i].a, roads[i].w);
            vis[i] = true;
            mst += roads[i].w;
            F[fb] = fa;
            ++cnt;
            if(cnt == n - 1) break;
        }
    }
    // 初始化
    sts[0][0] = INT_MIN;
    stm[0][0] = INT_MIN;
    sts[1][0] = INT_MIN;
    stm[1][0] = INT_MIN;
    // 如果调用dfs(1, 0) 需要初始化depth[0]
    depth[0] = -1;
    dfs(1, -1);
    ans = LONG_MAX;

    for(int i = 1; i <= m; ++i){
        if(vis[i]) continue;
        if(roads[i].a == roads[i].b) continue;
        int p = lca(roads[i].a, roads[i].b);

        int nw = max(query(roads[i].a, p, roads[i].w), query(roads[i].b, p, roads[i].w));
        //printf("lca(%d %d) = %d, nw = %d\n", roads[i].a, roads[i].b, p, nw);
        if(nw != INT_MIN)
            ans = min(ans, mst - nw + roads[i].w);
    }
    printf("%lld\n", ans);
    return 0;
}

4、Prim

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// https://www.luogu.com.cn/problem/P3366
using namespace std;
constexpr int maxn = 1000010;
using ll = long long int;
struct Edge {
    int next, to, w;
} edges[maxn * 2];
int heads[maxn], pos = 0;

void addEdge(int a, int b, int w) {
    edges[++pos].next = heads[a];
    edges[pos].to = b;
    edges[pos].w = w;
    heads[a] = pos;
}
bool vis[maxn];
int dist[maxn];
int main() {
    int n, m, a, b, w, cnt = 0;
    ll ans = 0;
    scanf("%d%d", &n, &m);
    memset(heads, -1, sizeof(heads));
    memset(vis, 0, sizeof(vis));
    memset(dist, 63, sizeof(dist));
    for (int i = 1; i <= m; ++i) {
        scanf("%d%d%d", &a, &b, &w);
        addEdge(a, b, w);
        addEdge(b, a, w);
    }
    using pi = pair<int, int>;
    priority_queue<pi, vector<pi>, greater<>> q;
    q.emplace(0, 1);
    while (!q.empty()) {
        pi p = q.top(); q.pop();
        if (vis[p.second]) continue;
        vis[p.second] = true;
        ++cnt;
        ans += p.first;
        //printf("to %d val %d\n", p.second, p.first);
        if (cnt == n) break;
        for (int i = heads[p.second]; ~i; i = edges[i].next) {
            int v = edges[i].to;
            int nw = edges[i].w;
            if (!vis[v] && dist[v] > nw) q.emplace(dist[v] = nw, v);
        }
    
    }
    if (cnt == n) printf("%lld\n", ans);
    else printf("orz\n");
    return 0;
}

发表于 更新于

一、常用模板

1、树上倍增

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// https://www.luogu.com.cn/problem/P3379
const int maxn = 500010;
using ll = long long int;
struct Edge {
    int next, to;
} edges[maxn * 2];
int heads[maxn], pos = 0;

void addEdge(int a, int b) {
    edges[++pos].next = heads[a];
    edges[pos].to = b;
    heads[a] = pos;
}
int st[maxn][20];
int depth[maxn];
void dfs(int u, int fa){
    depth[u] = fa == -1 ? 0 : depth[fa] + 1;
    st[u][0] = fa;
    for(int i = 1; ~st[u][i-1] && i < 20; ++i){
        st[u][i] = st[st[u][i-1]][i-1];
    }
    for(int i = heads[u]; ~i; i = edges[i].next){
        int v = edges[i].to;
        if(fa == v) continue;
        dfs(v, u);
    }
}
int lca(int u, int v){
    if(depth[u] < depth[v]) return lca(v, u);
    for(int i = 19; ~i; --i){
        if(~st[u][i] && depth[st[u][i]] >= depth[v]) u = st[u][i];
    }
    if(u == v) return u;
    for(int i = 19; ~i; --i){
        if(st[u][i] != st[v][i]) u = st[u][i], v = st[v][i];
    }
    return st[u][0];
}
int main(){
    int n, m, s, a, b;
    memset(heads, -1, sizeof(heads));
    memset(st, -1, sizeof(st));
    scanf("%d%d%d", &n, &m, &s);
    for(int i = 1; i < n; ++i){
        scanf("%d%d", &b, &a);
        if(a == b) continue;
        addEdge(a, b);
        addEdge(b, a);
    }
    dfs(s, -1);
    for(int i = 0; i < m; ++i){
        scanf("%d%d", &b, &a);
        printf("%d\n", lca(a, b));
    }
    return 0;
}

2、Tarjan

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const int maxn = 500010;
using ll = long long int;

struct Edge {
    int next, to;
} edges[maxn * 2];
int heads[maxn], pos = 0;
using pi = pair<int, int>;
vector<pi> qst[maxn];
int ans[maxn];
void addEdge(int a, int b) {
    edges[++pos].next = heads[a];
    edges[pos].to = b;
    heads[a] = pos;
}
int F[maxn];
bool vis[maxn];
int Find(int x){
    return x == F[x] ? x : F[x] = Find(F[x]);
}
void dfs(int u, int fa){
    F[u] = u;
    vis[u] = true;
    for(int i = heads[u]; ~i; i = edges[i].next){
        int v = edges[i].to;
        if(fa == v) continue;
        dfs(v, u);
    }
    for(auto &p : qst[u]){
        if(vis[p.first]) ans[p.second] = Find(p.first);
    }
    F[u] = fa;
}

int main(){
    int n, m, s, a, b;
    memset(heads, -1, sizeof(heads));
    memset(vis, 0, sizeof(vis));
    scanf("%d%d%d", &n, &m, &s);
    for(int i = 0; i < n; ++i) qst[i].clear();
    for(int i = 1; i < n; ++i){
        scanf("%d%d", &b, &a);
        if(a == b) continue;
        addEdge(a, b);
        addEdge(b, a);
    }

    for(int i = 0; i < m; ++i){
        scanf("%d%d", &b, &a);
        qst[a].emplace_back(b, i);
        qst[b].emplace_back(a, i);
    }
    dfs(s, s);
    for(int i = 0; i < m; ++i) printf("%d\n", ans[i]);
    return 0;
}

3、RMQ

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using namespace std;
const int maxn = 500010;
using ll = long long int;

struct Edge {
    int next, to;
} edges[maxn * 2];
int heads[maxn], pos = 0;
void addEdge(int a, int b) {
    edges[++pos].next = heads[a];
    edges[pos].to = b;
    heads[a] = pos;
}
int st[maxn * 2][20], idx = -1;
int loc[maxn], depth[maxn];
int Log2[maxn * 2];
void dfs(int u, int fa){
    depth[u] = fa == u ? 0 : depth[fa] + 1;
    st[++idx][0] = u;
    loc[u] = idx;
    for(int i = heads[u]; ~i; i = edges[i].next){
        int v = edges[i].to;
        if(fa == v) continue;
        dfs(v, u);
        st[++idx][0] = u;
    }
}
int minNode(int u, int v){
    return depth[u] < depth[v] ? u : v;
}
int query(int u, int v){
    u = loc[u], v = loc[v];
    if(u > v) swap(u, v);
    int k = Log2[v - u + 1];
    return minNode(st[u][k], st[v - (1 << k) + 1][k]);
}
int main(){
    int n, m, s, a, b;
    memset(heads, -1, sizeof(heads));
    Log2[1] = 0, Log2[2] = 1;
    for(int i = 3, len = maxn * 2; i < len; ++i) Log2[i] = Log2[i >> 1] + 1;
    scanf("%d%d%d", &n, &m, &s);
    for(int i = 1; i < n; ++i){
        scanf("%d%d", &b, &a);
        if(a == b) continue;
        addEdge(a, b);
        addEdge(b, a);
    }
    dfs(s, s);
    // dfn 欧拉序列长度
    int len = n * 2 - 1;
    for(int i = 1; i < 20; ++i){
        for(int j = 0; j < len; ++j){
            int next = j + (1 << (i - 1));
            if(next >= len) break;
            st[j][i] = minNode(st[j][i - 1], st[next][i - 1]);
        }
    }
        for(int i = 0; i < m; ++i){
        scanf("%d%d", &b, &a);
        printf("%d\n", query(a, b));
    }
    return 0;
}

发表于 更新于

一、数据结构

1、ST表

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template<typename T, typename F>
struct disjoint_sparse_table {
    int n;
    vector<int> log2;
    vector<vector<T>> mat;
    F func;
    disjoint_sparse_table(const vector<T>& arr, const F& f) : n(int(arr.size())), func(f) {
        build_log2();
        mat.push_back(arr);
        int k = log2[n];
        for (int i = 1; i <= k; ++i) {
            mat.emplace_back(n);
            for (int j = 0; j < n - (1 << (i - 1)); ++j) {
                mat[i][j] = func(mat[i - 1][j], mat[i - 1][j + (1 << (i - 1))]);
            }
        }
    }
    void build_log2() {
        log2.resize(n + 5, 0);
        log2[1] = 0;
        log2[2] = 1;
        for (int i = 3; i <= n; ++i) log2[i] = log2[i >> 1] + 1;
    }
    T query(int l, int r) {
        int k = log2[r - l + 1];
        return func(mat[k][l], mat[k][r - (1 << k) + 1]);
    }
};
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// https://vjudge.net/problem/POJ-3264
using namespace std;
const int maxn = 100010;
int FA[maxn][18];
int FB[maxn][18];
int Log2[maxn];

inline int read()
{
    int x=0,f=1;int ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
    return x*f;
}
int pre() {
    Log2[1] = 0;
    Log2[2] = 1;
    for(int i = 3; i < maxn; ++i) Log2[i] = Log2[i >> 1] + 1;
}
int Max(int a, int b) {
    return a > b ? a : b;
}
int Min(int a, int b){
    return a < b ? a: b;
}
int Query(int l, int r){
    int k = Log2[r - l + 1];
    int a = Max(FA[l][k], FA[r-(1<<k) + 1][k]);
    int b = Min(FB[l][k], FB[r-(1<<k) + 1][k]);
    return a - b;
}


int main() {
    int m, n, k;
    pre();
    m = read();
    n = read();
    k = Log2[m];
    for (int i = 0; i < m; ++i) FA[i][0] = FB[i][0]= read();
    for (int i = 1; i <= k; ++i) {
        for (int j = 0; j < m - (1 << (i - 1)); ++j)
            FA[j][i] = Max(FA[j][i - 1], FA[j + (1 << (i - 1))][i - 1]),
            FB[j][i] = Min(FB[j][i - 1], FB[j + (1 << (i - 1))][i - 1]);
    }
    for(int i = 0; i < n; ++i){
        if(i != 0) puts("");
        int l = read() - 1, r = read() - 1;
        printf("%d", Query(l, r));
    }
    return 0;
}

2、求TopK的数据结构

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// https://leetcode.cn/problems/find-number-of-coins-to-place-in-tree-nodes/
// 保存最小的N个数字
template<int N>
struct T120A {
    int arr[N];
    int cnt = 0;
    T120A() {
        for (int i = 0; i < N; ++i) arr[i] = INT_MAX;
    }
    void add(int x) {
        int i = 0, j = N - 2;
        while (i < N && x >= arr[i]) ++i;
        if (i == N) return;
        while (j >= i) arr[j + 1] = arr[j], --j;
        arr[i] = x;
        if (cnt < N) ++cnt;
       // for (int k = 0; k < cnt; ++k) printf("%2d ", arr[k]); puts("");
    }
};
// 保存最大的N个数字
template<int N>
struct T120B {
    int arr[N];
    int cnt = 0;
    T120B() {
        for (int i = 0; i < N; ++i) arr[i] = INT_MIN;
    }
    void add(int x) {
        int i = 0, j = N - 2;
        while (i < N && x <= arr[i]) ++i;
        if (i == N) return;
        while (j >= i) arr[j + 1] = arr[j], --j;
        arr[i] = x;
        if (cnt < N) ++cnt;
    }
};

3、二维前缀和

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// 不要使用C++的类 存在虚函数指针 可能会被卡常
// 根据情况流式处理数据减少数据拷贝
// https://leetcode.cn/problems/range-sum-query-2d-immutable/description/
void makePresum(vector<vector<int>>& arr) {
	int n = arr.size(), m = arr[0].size();
	for (int i = 1; i < m; ++i) arr[0][i] += arr[0][i - 1];
	for (int i = 1; i < n; ++i) {
		arr[i][0] += arr[i - 1][0];
		for (int j = 1; j < m; ++j)
			arr[i][j] = arr[i - 1][j] + arr[i][j - 1] - arr[i - 1][j - 1];
	}
}
int _getVal(vector<vector<int>>& arr, int a, int b) {
	if (a < 0 || b < 0) return 0;
	return arr[a][b];
}
int sumRegion(vector<vector<int>>& arr, int a, int b, int c, int d) {
	return _getVal(arr, c, d) - _getVal(arr, c, b - 1) - _getVal(arr, a - 1, d)
		+ _getVal(arr, a - 1, b - 1);
}
int main() 
{
	return 0;
}

3、快速幂

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// https://leetcode.cn/problems/count-ways-to-group-overlapping-ranges/description/
const int M = 1e9 + 7;
int fp(int x, int y){
    using ll = long long int;
    ll ans = 1;
    ll b = x;
    while(y){
        if(y & 1) ans = ans * b % M;
        b = b * b % M;
        y >>= 1;
    }
    return ans;
}

4、Dijkstra

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const int maxn = 100010;
struct Edge {
    int next, to, w;
} edges[maxn * 2];
int heads[maxn], pos = 0;

void addEdge(int a, int b, int w) {
    edges[++pos].next = heads[a];
    edges[pos].to = b;
    edges[pos].w = w;
    heads[a] = pos;
}
// 从0开始
bool vis[maxn];
int dist[maxn];
void dijkstra(int s, int n) {
    memset(vis, 0, sizeof(vis));
    memset(dist, 63, sizeof(dist));
    dist[s] = 0;
    using pi = pair<int, int>;
    priority_queue<pi, vector<pi>, greater<>> q;
    q.emplace(0, s);
    while (!q.empty()) {
        pi t = q.top(); q.pop();
        if (vis[t.second]) continue;
        vis[t.second] = true;
        for (int i = heads[t.second]; ~i; i = edges[i].next) {
            int v = edges[i].to, w = edges[i].w;
            if (!vis[v] && w + t.first < dist[v]) q.emplace(dist[v] = w + t.first, v);
        }
    }
}

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