hihocoder 1378 网络流二·最大流最小割定理

题目链接：网络流二·最大流最小割定理

题目大意：给你一张有向图，算最小割，最小割集的点数个数和个数编号

题目思路：跑一边最大流可以得到最小割，这个是显而易见的，要找最小割集，我们只需要最后再跑一遍BFS就好了，这个时候一定是找不到增广路的，所以这时候访问到的点就是最小割集了，记录一下就好了

#include <map>#include <set>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;const int MAXN = 1000+100;const int MAXM = 1000000+10;const int INF = 1000000+10;struct Edge{    int from, to, cap, flow, next;}edge[MAXM];int dist[MAXN], vis[MAXN], cur[MAXN], top, head[MAXN];int n, m;int sumflow;int sink;int num;bool flag;void init(){    top = 0;    memset(head, -1, sizeof(head));}void addedge(int u, int v, int w){    Edge E1 = {u, v, w, 0, head[u]};    edge[top] = E1;    head[u] = top++;    Edge E2 = {v, u, 0, 0, head[v]};    edge[top] = E2;    head[v] = top++;}bool BFS(int start, int End){    queue<int> Q;    memset(dist, -1, sizeof(dist));    memset(vis, 0, sizeof(vis));    while(!Q.empty()) Q.pop();    Q.push(start);    vis[start] = 1;    dist[start] = 0;    if(flag) num++;    while(!Q.empty()){        int u = Q.front();        Q.pop();        for(int i = head[u]; i != -1; i = edge[i].next){            Edge E = edge[i];            if(!vis[E.to] && E.cap > E.flow){                vis[E.to] = 1;                dist[E.to] = dist[u] + 1;                if(!flag&&E.to == End) return true;                if(flag) num++;                Q.push(E.to);            }        }    }    return false;}int DFS(int x, int a, int End){    if(x == End || a == 0) return a;    int flow = 0, f;    for(int& i = cur[x]; i != -1; i = edge[i].next){        Edge& E = edge[i];        if(dist[E.to] == dist[x]+1 && (f = DFS(E.to, min(a, E.cap-E.flow), End)) > 0){            E.flow += f;            edge[i^1].flow -= f;            flow += f;            a -= f;            if(a == 0) break;        }    }    return flow;}int Dinic(int start, int End){    int flow = 0;    while(BFS(start, End)){        memcpy(cur, head, sizeof(head));        flow += DFS(start, INF, End);    }    flag = true;    BFS(1,n);    return flow;}int main(){    int T,u,v,cost;    init();    num = 0;    scanf("%d%d",&n,&m);    while(m--){        scanf("%d%d%d",&u,&v,&cost);        addedge(u,v,cost);    }    printf("%d ",Dinic(1,n));    printf("%d\n",num);    for(int i = 1;i <= n;i++){        if(vis[i]) printf("%d ",i);    }    return 0;}