AOJ 0189 Convenient Location （Floyd）

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题意：

求某一个办公室到其他所有办公室的总距离最短办公室数不超过10

输入：

多组输入，每组第一行为n (1 ≤ n ≤ 45)，接下来n行是 (x, y, d)，x到y的距离是d
输出：

办公室号和最短距离

Floyd水题 - -

AC代码如下：

////  AOJ 0189 Convenient Location////  Created by TaoSama on 2015-03-20//  Copyright (c) 2015 TaoSama. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>#define CLR(x,y) memset(x, y, sizeof(x))using namespace std;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int N = 1e5 + 10;int n, dp[15][15];int main() {#ifdef LOCALfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endifios_base::sync_with_stdio(0);while(cin >> n && n) {int V = 0;memset(dp, 0x3f, sizeof dp);for(int i = 1; i <= n; ++i) {int x, y, v; cin >> x >> y >> v;dp[x][y] = dp[y][x] = v;V = max(V, max(x, y));}for(int k = 0; k <= V; ++k)for(int i = 0; i <= V; ++i)for(int j = 0; j <= V; ++j)dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);int ans = INF, loc;for(int i = 0; i <= V; ++i) {int t = 0;for(int j = 0; j <= V; ++j) {if(i == j) continue;t += dp[i][j];}if(t < ans) ans = t, loc = i;}cout << loc << ' ' << ans << endl;}return 0;}

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