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这题的思路还是比较简单,用弗洛伊德算法打表后,枚举来找到最小值
代码如下 注意最后判断时候的语句 在这里错误了很多次
# include<iostream> # include<algorithm> using namespace std; int p[105][105]; const int INF = 99999999; int n; void floyd() { for (int k = 1; k <= n; k++) for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) p[i][j] = min(p[i][j],p[i][k]+p[k][j]); } int main() { int m, e, t; while (cin >> n) { if (n == 0) break; for (int i = 1; i <= n; i++) for (int k = 1; k <= n; k++) { if (i == k) p[i][k] = 0; else p[i][k] = INF; } for (int i = 1; i <= n; i++) { cin >> m; if (m) { for (int k = 1; k <= m; k++) { cin >> e >> t; p[i][e] = t; } } } floyd(); int mymin = INF; bool flag = false; int s,j; for (int i = 1; i <= n; i++) { int mymax = -INF;//注意这个初始化写在循环里 for ( j = 1; j <= n; j++) { if (i != j) { if (p[i][j] == INF) break; mymax = max(mymax, p[i][j]); } } if (j == n + 1) flag = true; else continue; if (mymin > mymax) { mymin = mymax; s = i; } //上面这一部分的判断条件要想清楚 } if (flag) cout << s << " " << mymin << endl; else cout << "disjoint" << endl; } return 0; }
POJ1125-Stockbroker Grapevine Floyd算法多源最短路径
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原文地址:http://www.cnblogs.com/digitalhermit/p/5124055.html