最大流最小割——hihoCoder 1378

• 题目链接：
  https://hihocoder.com/problemset/problem/1378

• 分析：
  给出一个已经建立好的网络流图，这个图中的点被最小割分成两个部分S集合（包括源点）和T集合（包括汇点）

• 题解：
  先跑一边最大流，然后从源点开始BFS，如果搜到有边的容量还有剩余就加入S集合。因为最小割的边一定是满流的。

• AC代码：

/*************************************************************************    > File Name: test.cpp    > Author: Akira     > Mail: [e-mail]  ************************************************************************/#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cmath>#include <vector>#include <set>#include <list>#include <ctime>typedef long long LL;typedef unsigned long long ULL;typedef long double LD;#define MST(a,b) memset(a,b,sizeof(a))#define CLR(a) MST(a,0)#define Sqr(a) ((a)*(a))using namespace std;#define MaxN 666#define MaxM 41234#define INF 0x3f3f3f3f#define bug cout<<88888888<<endl;#define MIN(x,y) (x<y?x:y)#define MAX(x,y) (x>y?x:y)int N,M; //N个点，M条边struct Edge{    int u,v,next;    int flow;}edge[MaxM];   //最大边数，一般都是实际边数的2倍甚至以上int head[MaxN];int cont;void init(){        //记得init    cont = 0;    MST(head, -1);}void add(int u, int v, int flow){    edge[cont].u = u;    edge[cont].v = v;    edge[cont].flow = flow;    edge[cont].next = head[u];    head[u] = cont++;}void Add(int u, int v, int flow){    //建正反两条边，反向流量为0    add(u, v, flow);    add(v, u, 0);}int dis[MaxN];int num[MaxN];int cur[MaxN];int pre[MaxN];set<int> ans;bool flag;void BFS(int source,int sink){    queue<int>q;    CLR(num);    MST(dis,-1);    q.push(sink);    dis[sink]=0;    num[0]=1;    if(flag) ans.insert(sink);    while(!q.empty())    {        int u=q.front();        q.pop();        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v = edge[i].v;            if(flag)            {                if(dis[v] == -1 && edge[i].flow)                {                    dis[v] = dis[u] + 1;                    num[dis[v]]++;                    q.push(v);                    ans.insert(v);                }            }            else            {                if(dis[v] == -1)                {                    dis[v] = dis[u] + 1;                    num[dis[v]]++;                    q.push(v);                }            }        }    }}int ISAP(int source,int sink,int n)      //从源点到汇点，n为总点数，返回最大流{    memcpy(cur,head,sizeof(cur));    int flow=0, u = pre[source] = source;    BFS( source,sink);    while( dis[source] < n )    {        if(u == sink)        {            int df = INF, pos;            for(int i =source;i != sink;i = edge[cur[i]].v)            {                if(df > edge[cur[i]].flow)                {                    df = edge[cur[i]].flow;                    pos = i;                }            }            for(int i = source;i != sink;i = edge[cur[i]].v)            {                edge[cur[i]].flow -= df;                edge[cur[i]^1].flow += df;            }            flow += df;            //cout << flow << endl;            u = pos;        }        int st;        for(st = cur[u];st != -1;st = edge[st].next)        {            if(dis[edge[st].v] + 1 == dis[u] && edge[st].flow)            {                break;            }        }        if(st != -1)        {            cur[u] = st;            pre[edge[st].v] = u;            u = edge[st].v;        }        else        {            if( (--num[dis[u]])==0 ) break;            int mind = n;            for(int id = head[u];id != -1;id = edge[id].next)            {                if(mind > dis[edge[id].v] && edge[id].flow != 0)                {                    cur[u] = id;                    mind = dis[edge[id].v];                }            }            dis[u] = mind+1;            num[dis[u]]++;            if(u!=source)            u = pre[u];        }    }    return flow;}int u,v,w;int main(){    while( ~scanf("%d%d", &N, &M))    {        flag = false;        init();        ans.clear();        for(int i=0;i<M;i++)        {            scanf("%d%d%d", &u, &v, &w);            Add(u,v,w);        }        int S = 1, T = N;        cout << ISAP(1,N,N);        flag=true;        BFS(N,1);        printf(" %d\n", ans.size());        for(int x:ans)printf("%d ",x);        printf("\n");    }    //system("pause");}