网络流 dinic算法

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define pb push_back

const int N = 1e4 + 10;
const int M = 1e5 + 10;
const int INF = 2e9;
struct edge{
    int to, nxt, flow;
}e[M << 1];
int head[N], d[N], cur[N];
queue<int > que;
int n, m, s, t, tot;

//dinic算法 复杂度O(n^2*m) 二部图时O(n^0.5*m)
//bfs分层  增广路只能从上一层走到下一层
//dfs求流  得到该增广路的最小允许流

inline void add(int u, int v, int flow){
    e[tot].to = v; e[tot].flow = flow;
    e[tot].nxt = head[u]; head[u] = tot++;
    e[tot].to = u; e[tot].flow = 0;
    e[tot].nxt = head[v]; head[v] = tot++;
}

inline void init(){
    for(int i = 1; i <= n; ++i) cur[i] = head[i];
    while(!que.empty()) que.pop();
    for(int i = 1; i <= n; ++i) d[i] = 0;
}

bool bfs(int s, int t){
    d[s] = 1;
    que.push(s);
    int to, flow;
    while(!que.empty()){
        int now = que.front();
        que.pop();
        for(int o = head[now]; ~o; o = e[o].nxt){
            to = e[o].to;
            flow = e[o].flow;
            if(!d[to] && flow){
                d[to] = d[now] + 1;
                que.push(to);
            }
        }
    }
    if(d[t]) return true;
    else return false;
}

int dfs(int now, int t, int F){
    if(now == t) return F;

    int sum = 0;
    int to, flow, tmp;
    //当前弧优化，因为是dfs，如果遍历了当前边，当它使用之后，之后再用到这个边很大概率效果也是和之前一样，减少了一些时间开销
    for(int& o = cur[now]; ~o; o = e[o].nxt){
        to = e[o].to;
        flow = e[o].flow;
        if(flow && d[to] == d[now] + 1){
            tmp = dfs(to, t, min(F - sum, flow));

            //没有流  则不更新
            if(!tmp) continue;
            e[o].flow -= tmp; e[o^1].flow += tmp;
            sum += tmp;
            if(sum == F) return sum;
        }
    }

    return sum;
}

void dinic(int s,int& sum_flow){
    for(int i = 1; i <= n; ++i) cur[i] = head[i];
    while(bfs(s, t)){
        sum_flow += dfs(s, t, INF);
        init();
    }
}

void solve(){

    while(~scanf("%d%d%d%d", &n, &m, &s, &t)){
        int u, v, w;
        for(int i = 1; i <= n; ++i) head[i] = -1; tot = 0;
        for(int i = 1; i <= m; ++i){
            scanf("%d%d%d", &u, &v, &w);
            add(u, v, w);
        }
        int sum_flow = 0;
        dinic(s, sum_flow);
        //printf("sum_flow = %d
", sum_flow);
        printf("%d
", sum_flow);
    }
}

int main(){

    solve();

    return 0;
}