poj 3463 Sightseeing( 最短路与次短路)

http://poj.org/problem?id=3463

Sightseeing

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6420		Accepted: 2270

Description

Tour operator Your Personal Holiday organises guided bus trips across the Benelux. Every day the bus moves from one city S to another city F. On this way, the tourists in the bus can see the sights alongside the route travelled. Moreover, the bus makes a number of stops (zero or more) at some beautiful cities, where the tourists get out to see the local sights.

Different groups of tourists may have different preferences for the sights they want to see, and thus for the route to be taken from S to F. Therefore, Your Personal Holiday wants to offer its clients a choice from many different routes. As hotels have been booked in advance, the starting city S and the final city F, though, are fixed. Two routes from S to F are considered different if there is at least one road from a city A to a city B which is part of one route, but not of the other route.

There is a restriction on the routes that the tourists may choose from. To leave enough time for the sightseeing at the stops (and to avoid using too much fuel), the bus has to take a short route from S to F. It has to be either a route with minimal distance, or a route which is one distance unit longer than the minimal distance. Indeed, by allowing routes that are one distance unit longer, the tourists may have more choice than by restricting them to exactly the minimal routes. This enhances the impression of a personal holiday.

For example, for the above road map, there are two minimal routes from S = 1 to F = 5: 1 → 2 → 5 and 1 → 3 → 5, both of length 6. There is one route that is one distance unit longer: 1 → 3 → 4 → 5, of length 7.

Now, given a (partial) road map of the Benelux and two cities S and F, tour operator Your Personal Holiday likes to know how many different routes it can offer to its clients, under the above restriction on the route length.

Input

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:

• One line with two integers N and M, separated by a single space, with 2 ≤ N ≤ 1,000 and 1 ≤ M ≤ 10, 000: the number of cities and the number of roads in the road map.

• M lines, each with three integers A, B and L, separated by single spaces, with 1 ≤ A, B ≤ N, A ≠ B and 1 ≤ L ≤ 1,000, describing a road from city A to city B with length L.

  The roads are unidirectional. Hence, if there is a road from A to B, then there is not necessarily also a road from B to A. There may be different roads from a city A to a city B.

• One line with two integers S and F, separated by a single space, with 1 ≤ S, F ≤ N and S ≠ F: the starting city and the final city of the route.

  There will be at least one route from S to F.

Output

For every test case in the input file, the output should contain a single number, on a single line: the number of routes of minimal length or one distance unit longer. Test cases are such, that this number is at most 10⁹ = 1,000,000,000.

Sample Input

2
5 8
1 2 3
1 3 2
1 4 5
2 3 1
2 5 3
3 4 2
3 5 4
4 5 3
1 5
5 6
2 3 1
3 2 1
3 1 10
4 5 2
5 2 7
5 2 7
4 1

Sample Output

3
2

Hint

The first test case above corresponds to the picture in the problem description.

Source

BAPC 2006 Qualification

 

【题解】:

求最短路与次短路，如果  最短路+1 == 次短路  ，输出 最短路与次短路的条数和 否则 输出 最短路的条数

　　更新状态时：
1）新值小于最短路径长：更新最短路径长，计数；次短路径长，计数
2）新值等于最短路径长：更新最短路径计数
3）新值大于最短路径长，小于次短路径长：更新次短路径长，计数
4）新值等于次短路径长：更新次短路径计数

【注意】：不能用邻接矩阵存图，因为可能出现重复的路，应该用邻接链表

 

【code】：

 /**
 Judge Status:Accepted    Memory:1028K
 Time:94MS    Language:G++
 Code Length:2335B  Author:cj
 */

 #include<iostream>
 #include<stdio.h>
 #include<algorithm>
 #include<string.h>
 #include<vector>

 using namespace std;

 #define N 1010
 #define INF 1000000000

 struct Nod
 {
     int v,w;  //v 和 u->v 的权值 w
 }nd;

 vector<Nod> G[N];  //邻接链表保存图

 int dis[][N],visit[][N],cnt[][N];
 int n;

 int Dijkstra(int u,int t)
 {
     int i,j;
     for(i=;i<=n;i++)
     {
         dis[][i]=dis[][i]=INF;  //各种初始化
     }
     memset(cnt,,sizeof(cnt));
     memset(visit,,sizeof(visit));
     dis[][u] = ;
     cnt[][u] = ;  //到达起点的最短路有一条本身
     int flag = ;
     for(i=;i<=*n;i++)  //循环2*n次，每次求最短路 或者 次短路
     {
         int mins = INF;
         for(j=;j<=n;j++)
         {
             if(!visit[][j]&&dis[][j]<mins) //每次找一个最小
             {
                 flag = ;
                 u = j;
                 mins = dis[][j];
             }
             else if(!visit[][j]&&dis[][j]<mins) //每次找一个最小
             {
                 u = j;
                 flag = ;
                 mins = dis[][j];
             }
         }
         if(mins==INF)    break;  //没有找到最小就break掉
         visit[flag][u] = ;  //每次找一个最小
         for(j=;j<G[u].size();j++)
         {
             int v = G[u][j].v;
             int w = G[u][j].w;
             if(dis[][v]>mins+w)  //小于最短路
             {
                 dis[][v] = dis[][v];
                 cnt[][v] = cnt[][v];
                 dis[][v] = mins + w;
                 cnt[][v] = cnt[flag][u];
             }
             else if(dis[][v]==mins+w)  //等于最短路
             {
                 cnt[][v]+=cnt[flag][u];
             }
             else if(dis[][v]>mins+w)  //小于次短路
             {
                 dis[][v] = mins+w;
                 cnt[][v] = cnt[flag][u];
             }
             else if(dis[][v]==mins+w)  //等于次短路
             {
                 cnt[][v]+=cnt[flag][u];
             }
         }
     }
     int num = cnt[][t];
     if(dis[][t]+==dis[][t]) //如果次短路等于最短路加1
     {
         num+=cnt[][t];
     }
     return num;
 }

 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         int m;
         scanf("%d%d",&n,&m);
         int i,a,b,c;
         for(i=;i<=n;i++)
         {
             G[i].clear();
         }
         for(i=;i<m;i++)
         {
             scanf("%d%d%d",&a,&b,&c);
             nd.v = b;
             nd.w = c;
             G[a].push_back(nd);
         }
         scanf("%d%d",&a,&b);
         printf("%d\n",Dijkstra(a,b));
     }
     return ;
 }