POJ-3469-Dual Core CPU

来源：互联网发布：淘宝和别人互刷商品编辑：程序博客网时间：2024/05/21 11:33

题目不难，就是求最小割，用网络流做即可，

建图完毕后用Dinic做的，卡超时了很久，唉~最后找到一种新的优化方法才勉强过的

代码：

#include<cstdio>#include<cstring>#include<iostream>using namespace std;const int maxn=20100;const int maxm=500100;const int inf=1<<29;int n,m,st,des,e,pnt[maxm],nxt[maxm],cost[maxm],head[maxn],level[maxn],q[maxn];void AddEdge(int u,int v,int c1,int c2){    pnt[e]=v;nxt[e]=head[u];cost[e]=c1;head[u]=e++;    pnt[e]=u;nxt[e]=head[v];cost[e]=c2;head[v]=e++;}void read(){    st=0,des=n+1,e=0;    memset(head,-1,sizeof(head));    for(int i=1;i<=n;i++)    {int a,b;scanf("%d%d",&a,&b);AddEdge(st,i,a,0);AddEdge(i,des,b,0);    }    for(int i=1;i<=m;i++)    {int u,v,c;scanf("%d%d%d",&u,&v,&c);AddEdge(u,v,c,c);    }}bool BFS(){    memset(level,0,sizeof(level));    level[st]=1;    int pre=0,last=1;    q[pre]=st;    while(pre<last)    {if(q[pre]==des)    return true;for(int i=head[q[pre]];i!=-1;i=nxt[i])    if(cost[i]&&!level[pnt[i]])    {level[pnt[i]]=level[q[pre]]+1;q[last++]=pnt[i];    }    pre++;    }    return level[des];}int DFS(int u,int sum){    if(u==des||!sum)return sum;    int ans=0;    for(int i=head[u],t;i!=-1&&sum>ans;i=nxt[i])if(cost[i]&&level[pnt[i]]==level[u]+1){    int t=DFS(pnt[i],min(sum-ans,cost[i]));    if(!t)    {level[pnt[i]]=0;continue;    }    cost[i]-=t;    cost[i^1]+=t;    ans+=t;}    return ans;}void Dinic(){    int ans=0;    while(BFS())ans+=DFS(st,inf);    printf("%d\n",ans);}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {read();Dinic();    }    return 0;}