UVA558 LA5579 Wormholes【Floyd算法】

In the year 2163, wormholes were discovered. A wormhole is a subspace tunnel through space and time connecting two star systems. Wormholes have a few peculiar properties:
• Wormholes are one-way only.
• The time it takes to travel through a wormhole is negligible.
• A wormhole has two end points, each situated in a star system.
• A star system may have more than one wormhole end point within its boundaries.
• For some unknown reason, starting from our solar system, it is always possible to end up in any star system by following a sequence of wormholes (maybe Earth is the centre of the universe).
• Between any pair of star systems, there is at most one wormhole in either direction.
• There are no wormholes with both end points in the same star system.
    All wormholes have a constant time difference between their end points. For example, a specific wormhole may cause the person travelling through it to end up 15 years in the future. Another wormhole may cause the person to end up 42 years in the past.
    A brilliant physicist, living on earth, wants to use wormholes to study the Big Bang. Since warp drive has not been invented yet, it is not possible for her to travel from one star system to another one directly. This can be done using wormholes, of course.
    The scientist wants to reach a cycle of wormholes somewhere in the universe that causes her to end up in the past. By travelling along this cycle a lot of times, the scientist is able to go back as far in time as necessary to reach the beginning of the universe and see the Big Bang with her own eyes. Write a program to find out whether such a cycle exists.
Input
The input file starts with a line containing the number of cases c to be analysed. Each case starts with a line with two numbers n and m. These indicate the number of star systems (1 ≤ n ≤ 1000) and the number of wormholes (0 ≤ m ≤ 2000). The star systems are numbered from 0 (our solar system) through n − 1 . For each wormhole a line containing three integer numbers x, y and t is given. These numbers indicate that this wormhole allows someone to travel from the star system numbered x to the star system numbered y, thereby ending up t (−1000 ≤ t ≤ 1000) years in the future.
Output
The output consists of c lines, one line for each case, containing the word ‘possible’ if it is indeed possible to go back in time indefinitely, or ‘not possible’ if this is not possible with the given set of star systems and wormholes.
Sample Input
2
3 3
0 1 1000
1 2 15
2 1 -42
4 4
0 1 10
1 2 20
2 3 30
3 0 -60
Sample Output
possible
not possible

问题链接：UVA558 LA5579 Wormholes
问题简述：（略）
问题分析：用Floyd算法解决，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA558 LA5579 Wormholes */#include <bits/stdc++.h>using namespace std;/* Floyd-Warshall算法：计算图中任意2点之间的最短距离* 复杂度：O(N×N×N)* 输入：n 全局变量，图结点数*      g 全局变量，邻接矩阵，g[i][j]表示结点i到j间的边距离* 输出：g 全局变量*/
const int INF = 0x3f3f3f3f;
const int N = 1000;
int g[N + 1][N + 1], n;void floyd()
{for(int k = 0; k < n; k++)for(int i = 0; i < n; i++)if(g[i][k] < INF)for(int j = 0; j < n; j++)if(g[k][j] < INF)g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
}int main()
{int t, m, x, y, time;scanf("%d", &t);while(t--) {scanf("%d%d", &n, &m);memset(g, INF, sizeof(g));for(int i = 0; i < m; i++) {scanf("%d%d%d", &x, &y, &time);g[x][y] = time;}floyd();int flag = 1;for(int i = 0; i < n && flag; i++)for(int j = 0; j < n; j++)if(g[i][j] < INF && g[j][i] < INF) {if(g[i][j] + g[j][i] < 0) {flag = 0;break;}}puts(flag ? "not possible" : "possible");}return 0;
}

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