[POJ1273][USACO4.2]Drainage Ditches (网络流最大流)

题意

网络流最大流模板

思路

EK也不会超时

所以说是一个数据比较水的模板题

但是POJ有点坑，多组数据，而且题目没给

哭得我AC率直掉

代码

用的朴素Dinic

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
#define N 200010
#define INF 0x3fffffff
inline int read() {
    int x = 0,f = 1;
    char ch = getchar();
    while(ch<'0'||ch>'9') { if(ch=='-') f = -1; ch = getchar(); }
    while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    return x * f;
}
struct point{
    int fr,to,val;
    int nxt;
}edge[N];
int head[N];
int cnt;
void add_edge(int x,int y,int z)
{
    edge[cnt].fr=x,edge[cnt].to=y,edge[cnt].val=z;
    edge[cnt].nxt=head[x],head[x]=cnt++;
    edge[cnt].fr=y,edge[cnt].to=x,edge[cnt].val=0;
    edge[cnt].nxt=head[y],head[y]=cnt++;
}
//我放弃链式前向星
int n,m,st,ed;
int deep[N];
int BFS()
{
    queue<int>q;
    memset(deep,0,sizeof(deep));
    q.push(st);
    deep[st]=1;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1;i=edge[i].nxt)
        {
            int tmp=edge[i].to;
            if(edge[i].val<=0 || deep[tmp]) continue;
            deep[tmp]=deep[u]+1;
            q.push(tmp);
        }
    }
    return deep[ed];
}
int ans;
int dfs(int u,int flow)//flow为到达终点最多能增广的值
{
    if(u==ed) return flow;
    int add=0;
    for(int i=head[u];i!=-1&&add<flow;i=edge[i].nxt)
    {
        int v=edge[i].to;
        if(deep[v]!=deep[u]+1) continue;
        if(!edge[i].val) continue;//剪枝
        int tmpadd=dfs(v,min(edge[i].val,flow-add));
        edge[i].val-=tmpadd;
        edge[i^1].val+=tmpadd;//sub
        add+=tmpadd;
    }
    return add;
}
void Dinic()
{
    while(BFS()) ans+=dfs(st,INF);
}
int main()
{
    while(scanf("%d%d",&n,&m)==2)
    {
        ans=0;
        st=1,ed=m;
        memset(edge,0,sizeof(edge));
        memset(head,-1,sizeof(head));
        for(int i=1,u,v,co;i<=n;i++)
        {
            u=read(),v=read(),co=read();
            add_edge(u,v,co);
        }
        Dinic();
        printf("%d
",ans);
    }
    return 0;
}