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[Swust OJ 412]--医院设置(floyd算法)

2015-06-26 16:06 465 查看
题目链接:http://acm.swust.edu.cn/problem/412/

Time limit(ms): 1000        Memory limit(kb): 65535

Description
设有一棵二叉树,如图:

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define inf 0x3f3f3f

int n, ptr[101], mpt[101][101], L, R;
void Floyd(){
for (int k = 1; k <= n; k++){
for (int i = 1; i <= n; i++){
for (int j = 1; j <= n; j++)
mpt[i][j] = min(mpt[i][j], mpt[i][k] + mpt[k][j]);
}
}
}
int main(){
int i, j, sum, tmp;
while (cin >> n){
sum = inf;
memset(mpt, inf, sizeof(mpt));
for (i = 1; i <= n; i++){
cin >> ptr[i] >> L >> R;
mpt[L][i] = mpt[i][L] = mpt[R][i] = mpt[i][R] = 1;
}
Floyd();
for (i = 1; i <= n; i++){
tmp = 0;
for (j = 1; j <= n; j++) {
if (i != j)
tmp += mpt[i][j] * ptr[j];
}
sum = min(sum, tmp);
}
cout << sum << endl;
}
return 0;
}


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