复习图--WuKong

2024-08-25 01:32

文章标签 复习 wukong

本文主要是介绍复习图--WuKong，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

E - WuKong

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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Description

Liyuan wanted to rewrite the famous book “Journey to the West” (“Xi You Ji” in Chinese pinyin). In the original book, the Monkey King Sun Wukong was trapped by the Buddha for 500 years, then he was rescued by Tang Monk, and began his journey to the west. Liyuan thought it is too brutal for the monkey, so he changed the story:

One day, Wukong left his home - Mountain of Flower and Fruit, to the Dragon King’s party, at the same time, Tang Monk left Baima Temple to the Lingyin Temple to deliver a lecture. They are both busy, so they will choose the shortest path. However, there may be several different shortest paths between two places. Now the Buddha wants them to encounter on the road. To increase the possibility of their meeting, the Buddha wants to arrange the two routes to make their common places as many as possible. Of course, the two routines should still be the shortest paths.

Unfortunately, the Buddha is not good at algorithm, so he ask you for help.

Input

There are several test cases in the input. The first line of each case contains the number of places N (1 <= N <= 300) and the number of roads M (1 <= M <= N*N), separated by a space. Then M lines follow, each of which contains three integers a b c, indicating there is a road between place a and b, whose length is c. Please note the roads are undirected. The last line contains four integers A B C D, separated by spaces, indicating the start and end points of Wukong, and the start and end points of Tang Monk respectively.

The input are ended with N=M=0, which should not be processed.

Output

Output one line for each case, indicating the maximum common points of the two shortest paths.

Sample Input

Sample Output

     3Hint: One possible arrangement is (1-2-3-4-6) for Wukong and (2-3-4) for Tang Monk. The number of common points are 3.

题目大意，n个点m条路，然后是每条路连接的点以及长度，然后是四个数，由a到b 和由c到d，问在这两个路均是最短路的条件下，最多有几个点可以重合

要求最多的点重合，也就是说要在相同的最短路内尽量用更多的点，定义数组dis[i][j]表示i到j的最短距离，定义mm[i][j]表示i到j的最短距离需要几个点。相同的最短距离要求最多的点，弗洛伊德算法，求解每个点的最短路和需要的点数，遍历路线（i，j）使得 dis[a][b] = dis[a][i] + dis[i][j] + dis[j][b] ;满足条件的路线选取更多的点。

#include <cstdio>
#include <cstring>
#define INF 0x3f3f3f3f
#include <algorithm>
using namespace std;
int dis[310][310] , mm[310][310] ;
int main()
{int i , j , k , n , m , a , b , c , d ;while(~scanf("%d %d", &n, &m)){if(n == 0 && m == 0)break;memset(dis,INF,sizeof(dis));memset(mm,0,sizeof(mm));while(m--){scanf("%d %d %d", &i, &j, &k);if( dis[i][j] > k ){dis[i][j] = dis[j][i] = k ;mm[i][j] = mm[j][i] = 1 ;}}scanf("%d %d %d %d", &a, &b, &c, &d);for(i = 1 ; i <= n ; i++){dis[i][i] = 0 ;mm[i][i] = 0 ;}for(k = 1 ; k <= n ; k++)for(i = 1 ; i <= n ; i++)for(j = 1 ; j <= n ; j++)if( dis[i][j] > dis[i][k] + dis[k][j] || ( dis[i][j] == dis[i][k] + dis[k][j] && mm[i][j] < mm[i][k] + mm[k][j]  ) ){dis[i][j] = dis[i][k] + dis[k][j] ;mm[i][j] = mm[i][k] + mm[k][j] ;}int ans = -1 ;for(i = 1 ; i <= n ; i++)for(j = 1 ; j <= n ; j++)if( mm[i][j] > ans && dis[a][b] == dis[a][i] + dis[i][j] + dis[j][b] && dis[c][d] == dis[c][i] + dis[i][j] + dis[j][d]  )ans = mm[i][j] ;printf("%d\n", ans+1);}return 0;
}

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