题目大意：给出一个无向图，问从1到n的最短路发生变化需要割掉最少花费的边权总值是多少。

思路：先要把所有最短路上的边搞出来，一个Floyd就可以解决，然后把所有在最短路上的边都加到最大流的图中，然后跑最小割就是答案。

CODE：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 510
#define MAXE 300000
#define INF 0x3f3f3f3f
#define S 1
#define T points
using namespace std;
#define min(a,b) ((a) < (b) ? (a):(b))

struct Edge{
	int x,y,len;
	int cost;
	
	Edge(int _,int __,int ___,int ____):x(_),y(__),len(___),cost(____) {}
	Edge() {}
}edge[MAXE];

int points,edges;
int map[MAX][MAX];

struct MaxFlow{
	int head[MAX],total;
	int _next[MAXE << 1],aim[MAXE << 1],flow[MAXE << 1];
	
	int f[MAX],deep[MAX];
	bool v[MAX];
	
	MaxFlow() {
		total = 1;
		memset(head,0,sizeof(head));
	}
	void Add(int x,int y,int f) {
		_next[++total] = head[x];
		aim[total] = y;
		flow[total] = f;
		head[x] = total;
	}
	void Insert(int x,int y,int f) {
		Add(x,y,f);
		Add(y,x,0);
	}
	bool BFS() {
		static queue<int> q;
		while(!q.empty())	q.pop();
		memset(f,0x3f,sizeof(f));
		memset(deep,0,sizeof(deep));
		deep[S] = 1;
		f[S] = 0;
		q.push(S);
		while(!q.empty()) {
			int x = q.front(); q.pop();
			v[x] = false;
			for(int i = head[x]; i; i = _next[i])
				if(!deep[aim[i]] && flow[i]) {
					deep[aim[i]] = deep[x] + 1;
					q.push(aim[i]);
					if(aim[i] == T)	return true;
				}
		}
		return false;
	}
	int Dinic(int x,int f) {
		if(x == T)	return f;
		int temp = f;
		for(int i = head[x]; i; i = _next[i])
			if(flow[i] && deep[aim[i]] == deep[x] + 1 && temp) {
				int away = Dinic(aim[i],min(flow[i],temp));
				if(!away)	deep[aim[i]] = 0;
				flow[i] -= away;
				flow[i^1] += away;
				temp -= away;
			}
		return f - temp;
	}
}solver;

int main()
{
	cin >> points >> edges;
	memset(map,0x3f,sizeof(map));
	for(int i = 1; i <= points; ++i)
		map[i][i] = 0;
	for(int x,y,z,t,i = 1; i <= edges; ++i) {
		scanf("%d%d%d%d",&x,&y,&z,&t);
		map[x][y] = map[y][x] = z;
		edge[i] = Edge(x,y,z,t);
	}
	for(int k = 1; k <= points; ++k)
		for(int i = 1; i <= points; ++i)
			for(int j = 1; j <= points; ++j)
				map[i][j] = min(map[i][j],map[i][k] + map[k][j]);
	cout << map[1][points] << endl;
	for(int i = 1; i <= edges; ++i) {
		if(map[1][edge[i].x] + map[edge[i].y][points] + edge[i].len == map[1][points])
			solver.Insert(edge[i].x,edge[i].y,edge[i].cost);
		if(map[1][edge[i].y] + map[edge[i].x][points] + edge[i].len == map[1][points])
			solver.Insert(edge[i].y,edge[i].x,edge[i].cost);
	}
	int max_flow = 0;
	while(solver.BFS())
		max_flow += solver.Dinic(S,INF);
	cout << max_flow << endl;
	return 0;
}

BZOJ 1266 AHOI 2006 上学路线route 最小割