luoguP1401 城市

https://www.luogu.org/problemnew/show/P1401

二分答案网络流判断是否可行即可

#include <bits/stdc++.h>
using namespace std;

template <typename T>
inline void read(T &f) {
    f = 0; T fu = 1; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') fu = -1; c = getchar();}
    while (c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15); c = getchar();}
    f *= fu;
}

const int N = 200 + 5, M = 4e4 + 5, INF = INT_MAX;

struct Edge {
    int u, v, next, cap, flow;
}G[M << 1];

struct ele {
    int u, v, val;
    bool operator < (const ele A) const {return val < A.val;}
}p[M];

int head[N], nowhead[N], d[N];
int tot, n, m, t, S, T;

inline void addedge(int u, int v, int cap) {
    G[++tot] = (Edge) {u, v, head[u], cap, 0}, head[u] = tot;
    G[++tot] = (Edge) {v, u, head[v], cap, 0}, head[v] = tot;
}

int bfs() {
    memset(d, 0, sizeof(d));
    queue <int> q;
    d[S] = 1; q.push(S);
    while(!q.empty()) {
        int u = q.front(); q.pop();
        for(int i = head[u]; i; i = G[i].next) {
            int v = G[i].v;
            if(d[v] == 0 && G[i].cap > G[i].flow) {
                d[v] = d[u] + 1;
                q.push(v);
            }
        }
    }
    return d[T];
}

int dfs(int u, int Flow) {
    if(u == T || Flow == 0) return Flow;
    int flow = 0, f;
    for(int &i = nowhead[u]; i; i = G[i].next) {
        int v = G[i].v;
        if(d[v] == d[u] + 1 && (f = dfs(v, min(Flow, G[i].cap - G[i].flow))) > 0) {
            flow += f, Flow -= f;
            G[i].flow += f, G[i ^ 1].flow -= f;
            if(!Flow) break;
        }
    }
    return flow;
}

int dinic() {
    int ans = 0;
    while(bfs()) {
        for(int i = 1; i <= n; i++) nowhead[i] = head[i];
        ans += dfs(S, INF);
    }
    return ans;
}

int main() {
    read(n); read(m); read(t);
    for(int i = 1; i <= m; i++) read(p[i].u), read(p[i].v), read(p[i].val);
    sort(p + 1, p + m + 1);
    int l = 1, r = m; S = 1; T = n;
    while(l < r) {
        int mid = (l + r) >> 1;
        memset(head, 0, sizeof(head)); tot = 1;
        for(int i = 1; i <= mid; i++) addedge(p[i].u, p[i].v, 1);
        if(dinic() < t) l = mid + 1;
        else r = mid;
    }
    printf("%d\n", p[l].val);
    return 0;
}