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Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 41510 Accepted Submission(s): 18133
1 #include <cstdio> 2 #include <iostream> 3 using namespace std; 4 5 const int INF = 0x3f3f3f3f; 6 7 int map[110][110]; 8 int i, j, k, n, m; 9 10 void Floyd() 11 { 12 for(k=1; k<=n; k++) 13 for(i=1; i<=n; i++) 14 for(j=1; j<=n; j++) 15 if(map[i][j] > map[i][k] + map[k][j]) 16 map[i][j] = map[i][k] + map[k][j]; 17 printf("%d\n", map[1][n]); 18 } 19 int main() 20 { 21 while(~scanf("%d %d", &n, &m)) 22 { 23 if(n==0 && m==0) 24 break; 25 for(i=1; i<=n; i++) 26 for(j=1; j<=n; j++) 27 map[i][j] = (i==j?0:INF); 28 int u, v, w; 29 for(i=1; i<=m; i++){ 30 scanf("%d %d %d", &u, &v, &w); 31 if(map[u][v] > w) 32 map[u][v]=map[v][u]=w; 33 } 34 Floyd(); 35 } 36 return 0; 37 }
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原文地址:http://www.cnblogs.com/fengshun/p/4691813.html