P4180 模板严格次小生成树[BJWC2010]（严格次小生成树）

Posted 2021-03-09 blog-fgy

题目链接

题意如题

做法

• 先做一遍最小生成树
• 枚举添加每一条非树边的情况，每一次构成一棵基环树，在环上找一条最长边（如果等于该非树边就用环上的严格次小边）
• 倍增LCA，倍增预处理的时候顺便维护严格次大值和最大值（注意细节）
• （如果是非严格次小生成树则只需要维护最大值即可）

代码

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
#define reg register
#define LL long long
using namespace std;

const int maxN = 100005, maxM = 300005;

inline int read() {
    int x = 0; char ch = getchar();
    while(!isdigit(ch)) ch = getchar();
    while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
    return x;
}
int N, M;
LL Ans = 1e18, MST;

struct Edge {
    int dis, nxt, to;
}e[maxM << 1]; int cnte = 1, head[maxN];
inline void add_Edge(int i, int j, int k) {
    e[++cnte].dis = k, e[cnte].nxt = head[i], e[cnte].to = j, head[i] = cnte;
}

int fa[maxN];
int find(int x) {return fa[x] == x ? x : fa[x] = find(fa[x]);}
void merge(int x, int y) {fa[find(x)] = find(y);}

struct EE {
    int u, v, w;
    bool is;
}E[maxM];
bool cmp(EE x, EE y) {
    return x.w < y.w;
}
void Init() {
    N = read(), M = read();
    for(reg int i = 1; i <= M; ++i) {
        E[i].u = read(), E[i].v = read(), E[i].w = read();
    }    
}
void Kruskal() {
    int cnt = 0;
    sort(E + 1, E + M + 1, cmp);
    for(int i = 1; i <= N; ++i) fa[i] = i;
    for(reg int i = 1; i <= M; ++i) {
        reg int u = E[i].u, v = E[i].v, w = E[i].w;
        if(u == v) {
            E[i].is = true;
            continue;
        }
        if(find(u) != find(v)) {
            add_Edge(u, v, w), add_Edge(v, u, w);
            merge(u, v), ++cnt,
            E[i].is = true,
            MST += w;
        }
        if(cnt == N - 1) break;
    }    
}
int g[maxN][30], dep[maxN], f[maxN][30], h[maxN][30];
//g数组是最大值，h数组是严格次大值
void dfs(int u, int father) {
    for(reg int v, i = head[u]; i; i = e[i].nxt) {
        if((v = e[i].to) == father) continue;
        dep[v] = dep[u] + 1,
        f[v][0] = u, h[v][0] = g[v][0] = e[i].dis, 
        dfs(v, u);
    }
}
void Pre() {
    memset(h, -0x7f, sizeof(h)),
    memset(g, -0x7f, sizeof(g));
    dep[1] = 1, f[1][0] = 1;
    dfs(1, 0);
    for(int i = 1; i <= 21; ++i) {
        for(int u = 1; u <= N; ++u) {
            f[u][i] = f[f[u][i - 1]][i - 1],
　　　　　　　//注意一下处理
            g[u][i] = max(g[u][i - 1], g[f[u][i - 1]][i - 1]),
            h[u][i] = min(g[u][i - 1], g[f[u][i - 1]][i - 1]);
            if(i == 1) continue;
            if(g[u][i - 1] == g[f[u][i - 1]][i - 1]) {
                h[u][i] = max(h[u][i - 1], h[f[u][i - 1]][i - 1]);                
            }
            else {
                h[u][i] = max(h[u][i], h[u][i - 1]),
                h[u][i] = max(h[u][i], h[f[u][i - 1]][i - 1]);                
            } 
        }
    }
}
int LCA(int x, int y) {
    if(dep[x] < dep[y]) swap(x, y);
    for(int i = 20; i >= 0; --i) if(dep[f[x][i]] >= dep[y]) x = f[x][i];
    if(x == y) return x;
    for(int i = 20; i >= 0; --i) if(f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
    return f[x][0]; 
}
int Query(int down, int up, int now) {
    int ret = -1e18;
    for(int i = 20; i >= 0; --i) {
        if(dep[f[down][i]] >= dep[up]) {
            if(now != g[down][i]) ret = max(ret, g[down][i]);
            else ret = max(ret, h[down][i]);
            down = f[down][i];
        }
    }
    return ret;
}
void Solve() {
    Pre();
    for(int i = 1; i <= M; ++i) {
        if(E[i].is) continue;
        int x = E[i].u, y = E[i].v;
        int lca = LCA(x, y), k;
        k = Query(x, lca, E[i].w), k = max(k, Query(y, lca, E[i].w));
        if(k == E[i].w) continue;
        Ans = min(Ans, MST + (E[i].w - k));
    }
    printf("%lld
", Ans);
}
int main() {
    Init();
    Kruskal();
    Solve();
    return 0;
}