G

题意：有N个岛屿，M条路线，每条路都连接两个岛屿，并且每条路都有一个最大承载人数，现在想知道从最西边的岛到最东面的岛最多能有多少人过去（最西面和最东面的岛屿只有一个）。

分析：可以比较明显的看出来是一个最大流的问题，而且源点汇点也都给出了，可以用最基本的dinic解决，不过注意一些优化，都则会tle的，下面是dinic AC代码

============================================================================================================

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<queue>
#include<stack>
#include<algorithm>
#include<math.h>
using namespace std;

const int MAXN = 100005;
const int oo = 1e9+7;

struct Edge{int v, flow, next;}edge[MAXN<<1];
int Head[MAXN], cnt;
int layer[MAXN];

void InIt(int N)
{
    cnt = 0;
    memset(Head, -1, sizeof(Head));
}
void AddEdge(int u, int v, int flow)
{
    edge[cnt].v = v;
    edge[cnt].flow = flow;
    edge[cnt].next = Head[u];
    Head[u] = cnt++;
}
bool bfs(int start, int End)
{
    queue<int>Q;Q.push(start);
    memset(layer, 0, sizeof(layer));
    layer[start] = 1;

    while(Q.size())
    {
        int u = Q.front();Q.pop();

        if(u == End)return true;

        for(int j=Head[u]; j!=-1; j=edge[j].next)
        {
            int v = edge[j].v;

            if(layer[v] == false && edge[j].flow)
            {
                layer[v] = layer[u] + 1;
                Q.push(v);
            }
        }
    }

    return false;
}
int dfs(int u, int MaxFlow, int End)
{
    if(u == End)return MaxFlow;

    int uflow = 0;

    for(int j=Head[u]; j!=-1; j=edge[j].next)
    {
        int v = edge[j].v;

        if(layer[v]-1 == layer[u] && edge[j].flow)
        {
            int flow = min(MaxFlow-uflow, edge[j].flow);
            flow = dfs(v, flow, End);

            edge[j].flow -= flow;
            edge[j^1].flow += flow;

            uflow += flow;

            if(uflow == MaxFlow)
                break;
        }
    }

    if(uflow == 0)///优化防止重搜，不优化会超时
        layer[u] = 0;

    return uflow;
}
int dinic(int start, int End, int N)
{
    int MaxFlow = 0;

    while(bfs(start, End) == true)
        MaxFlow += dfs(start, oo, End);

    return MaxFlow;
}
int main()
{
    int T;

    scanf("%d", &T);

    while(T--)
    {
        int i, N, M, x, y, east=-oo, west=oo, start, End;

        scanf("%d%d", &N, &M);

        InIt(N);

        for(i=1; i<=N; i++)
        {
            scanf("%d%d", &x, &y);

            if(west > x)
            {
                west = x;
                start = i;
            }
            if(east < x)
            {
                east = x;
                End = i;
            }
        }

        while(M--)
        {
            int u, v, flow;

            scanf("%d%d%d", &u, &v, &flow);
            AddEdge(u, v, flow);
            AddEdge(v, u, flow);
        }

        printf("%d
", dinic(start, End, N));
    }

    return 0;
}