fzu1844 Earthquake Damage (网络流 最大流最小割/ISAP）

Earthquake Damage

Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

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Description

Open Source Tools help earthquake researchers stay a step ahead. Many geological research facilities around the world use or are in the process of developing open source software and applications designed to interpret and share information with other researchers. For example, OpenSees is an open source software framework for developing apps that help understand what happens to structures during and after earthquakes to help engineers design stronger buildings. Researchers are also using OpenSees to understand the potential ill-effects of seismic activity on viaducts and bridges.

China has had an earthquake that has struck Sichuan Province on Monday 12 May 2008!

The earthquake has damaged some of the cities so that they are unpassable. Remarkably, after repairing by Chinese People's Liberation Army, all the driveways between cities were fixed.

As usual, Sichuan Province is modeled as a set of P (1 <= P <= 3,000) cities conveniently numbered 1..P which are connected by a set of C (1 <= C <= 20,000) non-directional driveways conveniently numbered 1..C. Driveway i connects city a_i and b_i (1 <= a_i <= P; 1 <= b_i <= P). Driveway might connect a_i to itself or perhaps might connect two cities more than once. The Crisis Center is located in city 1.

A total of N (1 <= N <= P) survivors (in different cities) sequentially contacts Crisis Center via moobile phone with an integer message report_j (2 <= report_j <= P) that indicates that city report_j is undamaged but that the calling survivor is unable to return to the Crisis Center from city report_j because he/she could not find a path that does not go through damaged city.

After all the survivors report in, determine the minimum number of cities that are damaged.

Input

Input consists of several testcases. The format of each case as follow:

Line 1: Three space-separated integers: P, C, and N

Lines 2..C+1: Line i+1 describes cowpath i with two integers: a_i and b_i

Lines C+2..C+N+1: Line C+1+j contains a single integer: report_j

Output

For each testcase, output a line with one number, the minimum number of damaged cities.

Sample Input

5 5 21 22 33 52 44 545

Sample Output

1

Hint

Only city 2 being damaged gives such a scenario.

#include <cstdio>#include <cstring>#include <algorithm>const long M=6100;const long NE=121000;const long inf=1<<30;struct ISAP{    long head[M],next[NE],to[NE],cap[NE],cnt;    long dist[M],gap[M],pre[M],cur[M],n;    void init_edge(long x){        n=x;        memset(head,-1,sizeof(head));        cnt=0;    }    void add_edge(long u,long v,long c){        next[cnt]=head[u]; to[cnt]=v; cap[cnt]=c; head[u]=cnt++;        next[cnt]=head[v]; to[cnt]=u; cap[cnt]=0; head[v]=cnt++;    }    long MaxFlow(long s,long t){        long maxflow=0,flow=inf,flag;        long u,v;        for (long i=0;i<=n;++i){            dist[i]=gap[i]=0;            cur[i]=head[i];        }        u=pre[s]=s; gap[0]=n;        while (dist[s]<n){            flag=0;            for (long &i=cur[u];i!=-1;i=next[i]){                v=to[i];                if (cap[i]>0 && dist[u]==dist[v]+1){                    flow=std::min(flow,cap[i]);                    pre[v]=u;                    u=v;                    flag=1;                     if (v==t){                        maxflow+=flow;                        while (u!=s){                            u=pre[u];                            cap[cur[u]]-=flow;                            cap[cur[u]^1]+=flow;                        }                        flow=inf;                    }                    break;                }            }            if (flag) continue;            long min_dist=n;            for (long i=head[u];i!=-1;i=next[i]){                v=to[i];                if (cap[i]>0 && dist[v]<min_dist){                    min_dist=dist[v];                    cur[u]=i;                }            }            if ((--gap[dist[u]])==0) break;            gap[dist[u]=min_dist+1]++;            u=pre[u];        }        return maxflow;    }}G;int main(){    long p,c,n,vis[M];    while (~scanf("%d%d%d",&p,&c,&n)){long s=1,t=2*p+1;        G.init_edge(t);        memset(vis,0,sizeof(vis));        while (c--){            long u,v;            scanf("%d%d",&u,&v);            G.add_edge(u+p,v,inf);            G.add_edge(v+p,u,inf);        }        while (n--){            long x;            scanf("%d",&x);            vis[x]=1;            G.add_edge(x,t,inf);        }G.add_edge(1,1+p,inf);        for (long i=2;i<=p;++i)            if (!vis[i]) G.add_edge(i,i+p,1);            else G.add_edge(i,i+p,inf);        printf("%d\n",G.MaxFlow(s,t));    }    return 0;}