HDU4280 Island Transport【网络流】

来源：互联网发布：python 数据类型typeof 编辑：程序博客网时间：2024/04/28 20:22

题意：从最左边的点运人到最右边的点最多能运多少人

思路：裸的的最大流，就是卡时间，用邻接表的Dinic

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<stdlib.h>#include<math.h>#include<vector>#include<list>#include<map>#include<stack>#include<queue>#include<algorithm>#include<numeric>#include<functional>using namespace std;typedef long long ll;const int maxn = 100005;struct Edge{int to,next,cap;}edge[maxn*3];int head[maxn],tot,d[maxn];void init(){memset(head,-1,sizeof head);tot = 0;}void add(int x,int y,int cap){edge[tot].to = y;edge[tot].cap = cap;edge[tot].next = head[x];head[x] = tot++;edge[tot].to = x;edge[tot].cap = cap;edge[tot].next = head[y];head[y] = tot++;}bool bfs(int s,int t){    queue<int> p;    memset(d,0,sizeof(d));    p.push(s);    d[s]=1;    while(!p.empty())    {        int top=p.front();        p.pop();        if(top==t)            return true;        for(int i=head[top]; i!=-1; i=edge[i].next)        {            if(!d[edge[i].to] && edge[i].cap>0)            {                d[edge[i].to]=d[top]+1;                p.push(edge[i].to);            }        }    }    return false;}int dfs(int now,int t,int maxf){    int f,res=0;    if(now==t)        return maxf;    for(int i=head[now]; i!=-1; i=edge[i].next)    {        if((d[edge[i].to]==d[now]+1)&&edge[i].cap>0)        {            f=dfs(edge[i].to,t,min(maxf-res,edge[i].cap));            edge[i].cap-=f;            edge[i^1].cap+=f;            res+=f;            if(res==maxf)                return res;        }    }    if(res==0)        d[now]=0;    return res;}int maxflow(int s,int t){    int res=0;    while(bfs(s,t))        res+=dfs(s,t,0x3f3f3f3f);    return res;}int main(void){int T,n,m;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);init();int l = 0x3f3f3f3f,r = -0x3f3f3f3f;int s,t;for(int i = 1; i <= n; i++){int a,b;scanf("%d%d",&a,&b);if(a < l){l = a;s = i;}if(a > r){r = a;t = i;}}for(int i = 0; i < m; i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);add(a,b,c);}int ans = maxflow(s,t);printf("%d\n",ans);}return 0;}

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