HDU3416 Marriage Match IV —— 最短路径 + 最大流

题目链接：https://vjudge.net/problem/HDU-3416

Marriage Match IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4710    Accepted Submission(s): 1412

Problem Description

Do not sincere non-interference。
Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls are in city B. Every time starvae can get to city B and make a data with a girl he likes. But there are two problems with it, one is starvae must get to B within least time, it's said that he must take a shortest path. Other is no road can be taken more than once. While the city starvae passed away can been taken more than once.

So, under a good RP, starvae may have many chances to get to city B. But he don't know how many chances at most he can make a data with the girl he likes . Could you help starvae?

 

Input

The first line is an integer T indicating the case number.(1<=T<=65)
For each case,there are two integer n and m in the first line ( 2<=n<=1000, 0<=m<=100000 ) ,n is the number of the city and m is the number of the roads.

Then follows m line ,each line have three integers a,b,c,(1<=a,b<=n,0<c<=1000)it means there is a road from a to b and it's distance is c, while there may have no road from b to a. There may have a road from a to a,but you can ignore it. If there are two roads from a to b, they are different.

At last is a line with two integer A and B(1<=A,B<=N,A!=B), means the number of city A and city B.
There may be some blank line between each case.

 

Output

Output a line with a integer, means the chances starvae can get at most.

 

Sample Input

3
7 8
1 2 1
1 3 1
2 4 1
3 4 1
4 5 1
4 6 1
5 7 1
6 7 1
1 7

6 7
1 2 1
2 3 1
1 3 3
3 4 1
3 5 1
4 6 1
5 6 1
1 6

2 2
1 2 1
1 2 2
1 2

 

Sample Output

2
1
1

 

Author

starvae@HDU

 

Source

HDOJ Monthly Contest – 2010.06.05

题意：

求有多少条最短路径，要求边不能够重叠（但题目可以有重边，即两个点之间可以有几条边）。

题解：

1.先用spfa算法求出最短路径，并且筛掉那些不在最短路径上的边。

2.构建网络流图：对于在最短路径上的边u-->v，在网络流图中构建一条边 u-->v，且边权为1，表明这条边只能被走一次。

3.以起点为源点，以终点为汇点，求出最大流，即为答案。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int mod = 1e9+;
 const int MAXM = 1e5+;
 const int MAXN = 1e3+;

 vector<pair<int,int> >son[MAXN];
 int dis1[MAXN], dis2[MAXN], dis[MAXN], inq[MAXN];
 void spfa(int start, int N)
 {
     memset(inq, , sizeof(inq));
     for(int i = ; i<=N; i++)
         dis[i] = INF;

     queue<int>Q;
     Q.push(start);
     inq[start] = ;
     dis[start] = ;
     while(!Q.empty())
     {
         int u = Q.front();
         Q.pop(); inq[u] = ;
         for(int i = ; i<son[u].size(); i++)
         {
             int v = son[u][i].first;
             int w = son[u][i].second;
             if(dis[v]>dis[u]+w)
             {
                 dis[v] = dis[u]+w;
                 if(!inq[v])
                 {
                     Q.push(v);
                     inq[v] = ;
                 }
             }
         }
     }
 }

 struct Edge
 {
     int to, next, cap, flow;
 }edge[MAXM<<];
 int tot, head[MAXN];
 int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];

 void init()
 {
     tot = ;
     memset(head, -, sizeof(head));
 }

 void add(int u, int v, int w)
 {
     edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = ;
     edge[tot].next = head[u]; head[u] = tot++;
     edge[tot].to = u; edge[tot].cap = ; edge[tot].flow = ;
     edge[tot].next = head[v]; head[v] = tot++;
 }

 int sap(int start, int end, int nodenum)
 {
     memset(dep, , sizeof(dep));
     memset(gap, , sizeof(gap));
     memcpy(cur, head, sizeof(head));
     int u = pre[start] = start, maxflow = ,aug = INF;
     gap[] = nodenum;
     while(dep[start]<nodenum)
     {
         loop:
         for(int i = cur[u]; i!=-; i = edge[i].next)
         {
             int v = edge[i].to;
             if(edge[i].cap-edge[i].flow && dep[u]==dep[v]+)
             {
                 aug = min(aug, edge[i].cap-edge[i].flow);
                 pre[v] = u;
                 cur[u] = i;
                 u = v;
                 if(v==end)
                 {
                     maxflow += aug;
                     for(u = pre[u]; v!=start; v = u,u = pre[u])
                     {
                         edge[cur[u]].flow += aug;
                         edge[cur[u]^].flow -= aug;
                     }
                     aug = INF;
                 }
                 goto loop;
             }
         }
         int mindis = nodenum;
         for(int i = head[u]; i!=-; i = edge[i].next)
         {
             int v=edge[i].to;
             if(edge[i].cap-edge[i].flow && mindis>dep[v])
             {
                 cur[u] = i;
                 mindis = dep[v];
             }
         }
         if((--gap[dep[u]])==)break;
         gap[dep[u]=mindis+]++;
         u = pre[u];
     }
     return maxflow;
 }

 int u[MAXM], v[MAXM], w[MAXM];
 int main()
 {
     int T, n, m;
     scanf("%d", &T);
     while(T--)
     {
         int start, end;
         scanf("%d%d", &n,&m);
         for(int i = ; i<=m; i++)
             scanf("%d%d%d", &u[i], &v[i], &w[i]);
         scanf("%d%d", &start, &end);

         for(int i = ; i<=n; i++) son[i].clear();
         for(int i = ; i<=m; i++) son[u[i]].push_back(make_pair(v[i], w[i]));
         spfa(start, n);
         memcpy(dis1, dis, sizeof(dis1));
         int distance = dis1[end];

         for(int i = ; i<=n; i++) son[i].clear();
         for(int i = ; i<=m; i++) son[v[i]].push_back(make_pair(u[i], w[i]));
         spfa(end, n);
         memcpy(dis2, dis, sizeof(dis2));

         init();
         for(int i = ; i<=m; i++)
             if(dis1[u[i]]+w[i]+dis2[v[i]]==distance)
                 add(u[i], v[i], );

         int ans = sap(start, end, n);
         printf("%d\n", ans);
     }
 }