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UVAL1277_Cops and Thieves

单源点汇点无向图,要阻隔某个点的流量,必须在一个点上消耗一定的价值,问你能否在消耗价值不超过k的前提下,阻隔源点到汇点的流量。

直接对于有权值的点拆点,拆后边容量即为点权。其余的点的容量无穷,最大流即可。

 

 

召唤代码君:

 

 

#include <iostream>#include <cstring>#include <cstdio>#define maxn 5555#define maxm 555555using namespace std;const int inf=11111;int to[maxm],c[maxm],next[maxm],first[maxn],edge;int a[maxn],d[maxn],tag[maxn],TAG=222;bool can[maxn];int Q[maxm],bot,top;int n,m,l,s,t;void _init(){    edge=-1;    for (int i=1; i<=n+n; i++) first[i]=-1;}void addedge(int U,int V,int W){    edge++;    to[edge]=V,c[edge]=W,next[edge]=first[U],first[U]=edge;    edge++;    to[edge]=U,c[edge]=0,next[edge]=first[V],first[V]=edge;}bool bfs(){    Q[bot=top=1]=t,tag[t]=++TAG,d[t]=0,can[t]=false;    while (bot<=top)    {        int cur=Q[bot++];        for (int i=first[cur]; i!=-1; i=next[i])            if (c[i^1] && tag[to[i]]!=TAG)            {                tag[to[i]]=TAG,d[to[i]]=d[cur]+1;                can[to[i]]=false,Q[++top]=to[i];                if (to[i]==s) return true;            }    }    return false;}int dfs(int cur,int num){    if (cur==t) return num;    int tmp=num,k;    for (int i=first[cur]; i!=-1; i=next[i])        if (c[i] && d[to[i]]==d[cur]-1 && tag[to[i]]==TAG && !can[to[i]])        {            k=dfs(to[i],min(c[i],num));            if (k) num-=k,c[i]-=k,c[i^1]+=k;            if (!num) break;        }    if (num) can[cur]=true;    return tmp-num;}int main(){    int U,V,Flow,K;    while (scanf("%d",&K)!=EOF)    {        scanf("%d%d%d%d",&n,&m,&s,&t);        for (int i=1; i<=n; i++) scanf("%d",&a[i]);        _init();        for (int i=1; i<=m; i++)        {            scanf("%d%d",&U,&V);            addedge(U+n,V,inf),addedge(V+n,U,inf);        }        if (s==t)        {            puts("NO");            continue;        }        s+=n;        for (int i=1; i<=n; i++) addedge(i,i+n,a[i]);        for (Flow=0; Flow<=K && bfs(); bfs()) Flow+=dfs(s,inf);        if (K>=Flow) puts("YES");            else puts("NO");    }    return 0;}

 

UVAL1277_Cops and Thieves