hdu4289 最大流最小割

来源：互联网发布：世界遗产地理杂志淘宝编辑：程序博客网时间：2024/06/07 11:22
最小割问题拆点每个城市拆成两个点两点之间边的权值为之前点的权值就行了
//拆点 将点的权值变为边的权值#include<algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<string>#include<stack>#include<queue>#include<cmath>#include<stack>#include<list>#include<map>#include<set>typedef long long ll;using namespace std;const int MAXN=500;//jiedian de zui da zhiconst int MAXM=100000;//bian de zui da zhi 这里不要开小的 不然显示的是tleconst int INF=0x3f3f3f3f;struct Node{    int from,to,next;    int cap;}edge[MAXM];int tol;int head[MAXN];int dep[MAXN];int gap[MAXN];void init()  //remember write it in main function{    tol=0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int w){    edge[tol].from=u;    edge[tol].to=v;    edge[tol].cap=w;    edge[tol].next=head[u];    head[u]=tol++;    edge[tol].from=v;    edge[tol].to=u;    edge[tol].cap=0;//wuxiangtu  this place change to w;    edge[tol].next=head[v];    head[v]=tol++;}void BFS(int start,int end){    memset(dep,-1,sizeof(dep));    memset(gap,0,sizeof(gap));    gap[0]=1;    int que[MAXN];    int front,rear;    front=rear=0;    dep[end]=0;    que[rear++]=end;    while(front!=rear)    {        int u=que[front++];        if(front==MAXN)front=0;        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].to;            if(dep[v]!=-1)continue;            que[rear++]=v;            if(rear==MAXN)rear=0;            dep[v]=dep[u]+1;            ++gap[dep[v]];        }    }}int SAP(int start,int end,int n) //n shi jiedian de zui da ge shu ,including source and sink{    int res=0;    BFS(start,end);    int cur[MAXN];    int S[MAXN];    int top=0;    memcpy(cur,head,sizeof(head));    int u=start;    int i;    while(dep[start]<n)    {        if(u==end)        {            int temp=INF;            int inser;            for(i=0;i<top;i++)                if(temp>edge[S[i]].cap)                {                    temp=edge[S[i]].cap;                    inser=i;                }            for(i=0;i<top;i++)            {                edge[S[i]].cap-=temp;                edge[S[i]^1].cap+=temp;            }            res+=temp;            top=inser;            u=edge[S[top]].from;        }        if(u!=end&&gap[dep[u]-1]==0)            break;        for(i=cur[u];i!=-1;i=edge[i].next)            if(edge[i].cap!=0&&dep[u]==dep[edge[i].to]+1)                break;        if(i!=-1)        {            cur[u]=i;            S[top++]=i;            u=edge[i].to;        }        else        {            int min=n;            for(i=head[u];i!=-1;i=edge[i].next)            {                if(edge[i].cap==0)continue;                if(min>dep[edge[i].to])                {                    min=dep[edge[i].to];                    cur[u]=i;                }            }            --gap[dep[u]];            dep[u]=min+1;            ++gap[dep[u]];            if(u!=start)u=edge[S[--top]].from;        }    }    return res;}int main(){    int i,j,k;    int m,n;    while(scanf("%d%d",&n,&m)!=EOF)    {        init();        int source,sink;        scanf("%d%d",&source,&sink);        for(i=1;i<=n;i++)        {            int t;            scanf("%d",&t);            addedge(i, i+n, t);        }        int x,y;        while(m--)        {            scanf("%d%d",&x,&y);            addedge(x+n, y, INF);//注意是双向边            addedge(y+n, x, INF);        }                int tt=SAP(source,sink+n,2*n);//一共2*n个点        printf("%d\n",tt);    }        return 0;   }
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