Double Shortest Paths 网络流

题目：

Problem D. Double Shortest Paths

Alice and Bob are walking in an ancient maze with a lot of caves and one-way passages connecting them. They want to go from cave 1 to cave n. All the passages are difficult to pass. Passages are too small for two people to walk through simultaneously, and crossing a passage can make it even more difficult to pass for the next person. We define di as the difficulty of crossing passage i for the first time, and ai as the additional difficulty for the second time (e.g. the second person's difficulty is di+ai).

Your task is to find two (possibly identical) routes for Alice and Bob, so that their total difficulty is minimized.

For example, in figure 1, the best solution is 1->2->4 for both Alice and Bob, but in figure 2, it's better to use 1->2->4 for Alice and 1->3->4 for Bob.

Input

There will be at most 200 test cases. Each case begins with two integers n, m (1<=n<=500, 1<=m<=2000), the number of caves and passages. Each of the following m lines contains four integers u, v, di and ai (1<=u,v<=n, 1<=di<=1000, 0<=ai<=1000). Note that there can be multiple passages connecting the same pair of caves, and even passages connecting a cave and itself.

Output

For each test case, print the case number and the minimal total difficulty.

Sample Input

Output for Sample Input

4 4

1 2 5 1

2 4 6 0

1 3 4 0

3 4 9 1

4 4

1 2 5 10

2 4 6 10

1 3 4 10

3 4 9 10

Case 1: 23

Case 2: 24

题目大意：

从一走到n求走两次最小费用，如果是第二次走到该点，费用会增加。

题目思路：

1、题目和uva 10806 Dijkstra, Dijkstra. 很像

2、第一次可以用dijlstra算法，然后网络流增流，然后用flow算法

3、增流的时候要给路径加上附加值

4、.增流不要删掉之前的路。

程序：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <queue>
#include <stack>
#include <map>
#include <cmath>
#include <string>
#define INF 0x3f3f3f3f
using namespace std;
struct street
{
int x,y,f,d;
} a[2100];
bool vv[1000];
int n,m;
int v[1000];
int dij();
int dfs(int);
int main()
{
int ci=1;
while(~scanf("%d%d",&n,&m))
{
memset(vv,0,sizeof(vv));
int ans=0;
for(int i=0; i<m; i++)//cout<<'#';
scanf("%d%d%d%d",&a[i].x,&a[i].y,&a[i].f,&a[i].d);
ans+=dij();
dfs(n);
ans+=dij();
printf("Case %d: %d\n",ci++,ans);
}
return 0;
}
int dij()
{
memset(v,INF,sizeof(v));
v[1]=0;
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
if(v[a[j].x]!=INF)
v[a[j].y]=min(v[a[j].y],v[a[j].x]+a[j].f);
return v
;
}
int dfs(int dian)
{
if(dian==1)
return 1;
for(int i=0; i<m; i++)
{
if(a[i].y==dian)
if(!vv[a[i].x])
//if(!vv[a[i].y])
if(v[a[i].y]-v[a[i].x]==a[i].f)
{
vv[a[i].x]=1;
int t=m;
m++;
a[t].x=a[i].y;
a[t].y=a[i].x;
a[t].f=-a[i].f;
a[i].f+=a[i].d;
if(dfs(a[i].x))
return 1;
m--;
a[i].f-=a[i].d;
vv[a[i].x]=0;
}
}
}