poj 1125 Stockbroker Grapevine 【最短路 Floyd】


[align=center]Stockbroker Grapevine[/align]

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 31794		Accepted: 17420

题目大意：给你n个人的联系情况，对任意一个人，求出这个人发消息到其他n-1个人的时间，得到n-1个时间中的最大值，n个最大值中的最小值就是所求。如果网络不通，那就输出disjoint。
本题用Floyd更简洁
已Accept代码【c++】

<table width="100%" class="a" bordercolor="#ffffff" border="1" cellspacing="0" cellpadding="0"><tbody><tr align="center"><td><a target=_blank href="http://poj.org/problem?id=1125"><span style="color:#0000ff;">1125</span></a></td><td><span style="color:blue;">Accepted</span></td><td>140K</td><td>0MS</td><td>C++</td><td>923B</td></tr></tbody></table>

#include <cstdio>
#include <cstring>
using namespace std;
int map[101][101];
int n, m;

void Floyd() {
int v = 0;
for(int k = 1; k <= n; k++)
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
map[i][j] = map[i][j] > map[i][k] +map[k][j] ? map[i][k] + map[k][j] : map[i][j];
int Min = 100000;
for(int i = 1; i <= n; i++) {
int Max = -1;
for(int j = 1; j <= n; j++)
if(Max < map[i][j])
Max = map[i][j];
if(Max < Min) {
Min = Max;
v = i;
}
}
if(Min < 100000)
printf("%d %d\n", v, Min);
else
printf("disjoint\n");
}

int main() {
int a, b;
while(scanf("%d", &n), n) {
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
if(i == j)
map[i][j] = 0;
else
map[i][j] = 100000;
}
}
for(int i = 1; i <= n; i++) {
scanf("%d", &m);
for(int j = 0; j < m; j++) {
scanf("%d %d", &a, &b);
map[i][a] = b;
}
}
Floyd();
}
return 0;
}