ZOJ 2788 Panic Room

最小割问题。

设一个超级源点s,s向每个有入侵者的房间连一条容量为inf的边.然后对于两个房间的门（a，b），如果把手在a,连一条从a到b容量为inf的边，连一条从b到a容量为1的边。求一遍从s到被保护房间的最大流即可

#include<stdio.h>#include<cstring>#include<math.h>#define inf 1000000#define min(a,b) ((a)>(b))?(b):(a)using namespace std;int a,b;int n,k,dist[100],gap[100],edgeHead[100];//双向边struct{    int v,cap,next,re;}edge[1000];void addEdge(int u,int v,int ca,int ac){    edge[k].v=v;    edge[k].cap=ca;    edge[k].next=edgeHead[u];    edge[k].re=k + 1;               //这个用来记录此边的反边    edgeHead[u]=k ++;    edge[k].v=u;    edge[k].cap=ac;                 //这里是双向边    edge[k].next=edgeHead[v];    edge[k].re=k - 1;    edgeHead[v]=k ++;}int dfs (int p , int limit){    if(p==b)return limit;    for(int i=edgeHead[p];i!=0;i=edge[i].next){        int v = edge[i].v;        if(dist[p]==(dist[v]+1) && edge[i].cap>0){            int t=dfs(v,min(limit , edge[i].cap));            if(t<0)return t;//没有增广路了            if(t>0)//更新流            {                edge[i].cap-=t;                edge[edge[i].re].cap+=t;                return t;            }        }    }    int tmp=n+1;    for(int i=edgeHead[p];i!=0;i=edge[i].next){        int v = edge[i].v;        if(edge[i].cap>0)            tmp=min(tmp,dist[v]+1);    }    if(--gap[dist[p]]==0 || dist[0]>n)return -1;//出现断层或回流已满    ++gap[dist[p]=tmp];    return 0;}int SAP(){    gap[0]=n+1;    int f = 0 , t=0;    while (~(t=dfs(0,inf))) f+=t;    return f;}int main(){int i,j,T,t;char in[10];scanf("%d",&T);for(t=1;t<=T;t++){memset(edgeHead,0,sizeof(edgeHead));memset(dist,0,sizeof(dist));memset (gap , 0 , sizeof(gap));k=1;scanf("%d %d",&n,&b);b++;for(i=0;i<n;i++){scanf("%s",in);if(in[0]=='I')addEdge(0,i+1,inf,0);int has;scanf("%d",&has);for(j=1;j<=has;j++){int v;scanf("%d",&v);v++;addEdge(i+1,v,inf,1);}}int k=SAP();if(k>=inf)printf("PANIC ROOM BREACH\n");elseprintf("%d\n",k);}}