Island Transport

In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the islands relies on the ships.
　　You have a transportation company there. Some routes are opened for passengers. Each route is a straight line connecting two different islands, and it is bidirectional. Within an hour, a route can transport a certain number of passengers in one direction. For safety, no two routes are cross or overlap and no routes will pass an island except the departing island and the arriving island. Each island can be treated as a point on the XY plane coordinate system. X coordinate increase from west to east, and Y coordinate increase from south to north.
　　The transport capacity is important to you. Suppose many passengers depart from the westernmost island and would like to arrive at the easternmost island, the maximum number of passengers arrive at the latter within every hour is the transport capacity. Please calculate it.

Input

　　The first line contains one integer T (1<=T<=20), the number of test cases.
　　Then T test cases follow. The first line of each test case contains two integers N and M (2<=N,M<=100000), the number of islands and the number of routes. Islands are number from 1 to N.
　　Then N lines follow. Each line contain two integers, the X and Y coordinate of an island. The K-th line in the N lines describes the island K. The absolute values of all the coordinates are no more than 100000.
　　Then M lines follow. Each line contains three integers I1, I2 (1<=I1,I2<=N) and C (1<=C<=10000) . It means there is a route connecting island I1 and island I2, and it can transport C passengers in one direction within an hour.
　　It is guaranteed that the routes obey the rules described above. There is only one island is westernmost and only one island is easternmost. No two islands would have the same coordinates. Each island can go to any other island by the routes.

Output

　　For each test case, output an integer in one line, the transport capacity.

Sample Input

Sample Output

9
6

解析：这是一道基础的最大流，题目要求的是求出从最左边到最右边通过的最大流，而且每条路是双向的，代码如下，这道题不知道我之前写的代码怎么错了，怎么改都改不对，难受，等以后有时间再看吧，或是哪位大佬知道，欢迎指点

这是正确的代码


#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=100005;
const int maxx=200005;
struct node
{int v,w,next;
}e[maxx];
int head[maxn],vis[maxn];
int tot,S,T;
void addEdge(int u,int v,int cap)
{e[tot].v=v,e[tot].w=cap,e[tot].next=head[u],head[u]=tot++;e[tot].v=u,e[tot].w=cap,e[tot].next=head[v],head[v]=tot++;
}
bool bfs()
{queue<int> q;memset(vis,-1,sizeof(vis));q.push(S);vis[S]=0;while(!q.empty()){int u=q.front();q.pop();for(int i=head[u];~i;i=e[i].next){int v=e[i].v;if(vis[v]==-1&&e[i].w){vis[v]=vis[u]+1;if(v==T)return true;q.push(v);}}}return false;
}
int dfs(int u,int f)
{if(u==T||!f)return f;int r=0;for(int i=head[u];~i;i=e[i].next){int v=e[i].v;if(vis[v]==vis[u]+1&&e[i].w){int d=dfs(v,min(f,e[i].w));if(d>0){e[i].w-=d;e[i^1].w+=d;r+=d;f-=d;if(!f)break;}}}if(!r)vis[u]=-1;return r;
}int Dinic()//然后直接调用这个即可
{int ans=0;while(bfs())ans+=dfs(S,INF);return ans;
}void init()//记得每次使用前初始化
{memset(head,-1,sizeof(head));tot=0;
}int main()
{int t;scanf("%d",&t);int x,y,w;int n,m;int maxI,minI;while(t--){init();scanf("%d%d",&n,&m);scanf("%d%d",&x,&y);S=T=1;maxI=minI=x;for(int i=2;i<=n;i++){scanf("%d%d",&x,&y);if(maxI<x)T=i,maxI=x;if(minI>x)S=i,minI=x;}while(m--){scanf("%d%d%d",&x,&y,&w);addEdge(x,y,w);}int ans=Dinic();printf("%d\n",ans);}return 0;
}

这是一直超时的代码wa的我难受

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
using namespace std;
const int inf=0x3f3f3f3f;
const int M=200010;
const int mm=100010;
struct node
{int v,w,next;
} e[M];
int first[mm],vis[mm];
int tot=0,head,tail;
int m,n;
void init()
{tot=0;memset(first,-1,sizeof(first));
}
void add_edge(int u,int v,int w)
{e[tot].v=v;e[tot].w=w;e[tot].next=first[u];first[u]=tot++;e[tot].v=u;e[tot].w=w;e[tot].next=first[v];first[v]=tot++;
}
int bfs()
{memset(vis,-1,sizeof(vis));vis[head]=0;queue<int>q;while(!q.empty())q.pop();q.push(head);while(!q.empty()){int u=q.front();q.pop();for(int i=first[u]; i!=-1; i=e[i].next){int v=e[i].v;if(e[i].w>0&&vis[v]==-1){vis[v]=vis[u]+1;q.push(v);}}}return vis[tail]!=-1;
}
int dfs(int u,int w1)
{if(u==tail)return w1;int f=0,ans=0;for(int i=first[u]; i!=-1; i=e[i].next){int v=e[i].v,w=e[i].w;if(w>0&&vis[v]==vis[u]+1&&(f=dfs(v,min(w1,w)))){e[i].w-=f;e[i^1].w+=f;ans+=f;w1-=f;if(!f)break;}}vis[u]=-1;return ans;
}
void Dinic()
{int ans=0;while(bfs()){ans+=dfs(head,inf);}printf("%d\n",ans);
}
int main()
{int t,x,y,z;scanf("%d",&t);while(t--){init();scanf("%d%d",&n,&m);int maxx=-100000,minn=100000;for(int i=1; i<=n; i++){scanf("%d%d",&x,&y);if(x>maxx)maxx=x,tail=i;if(x<minn)minn=x,head=i;}for(int i=1; i<=m; i++){scanf("%d%d%d",&x,&y,&z);add_edge(x,y,z);}Dinic();}return 0;
}

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