POJ 2485 Highways 【最小生成树Prim】

最新推荐文章于 2019-08-15 08:57:00 发布

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最新推荐文章于 2019-08-15 08:57:00 发布

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原文链接：http://www.cnblogs.com/cniwoq/p/6770867.html

本文探讨了一种经典的图论问题——最小生成树的最大边问题，并提供了一个具体的代码实现案例。该问题要求在确保所有节点连通的前提下，找到最小生成树上最长边的最小可能长度。

Highways

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28397		Accepted: 12942

Description

The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public highways. So the traffic is difficult in Flatopia. The Flatopian government is aware of this problem. They're planning to build some highways so that it will be possible to drive between any pair of towns without leaving the highway system.

Flatopian towns are numbered from 1 to N. Each highway connects exactly two towns. All highways follow straight lines. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways.

The Flatopian government wants to minimize the length of the longest highway to be built. However, they want to guarantee that every town is highway-reachable from every other town.

Input

The first line of input is an integer T, which tells how many test cases followed.
The first line of each case is an integer N (3 <= N <= 500), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 65536]) between village i and village j. There is an empty line after each test case.

Output

For each test case, you should output a line contains an integer, which is the length of the longest road to be built such that all the villages are connected, and this value is minimum.

Sample Input

Sample Output

Hint

Huge input,scanf is recommended.

Source

POJ Contest,Author:Mathematica@ZSU

题意是求最小生成树上的最大的连接距离，题中直接给了矩阵，而且Prim算法更新最大的距离比较方便。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 1005
using namespace std;
const int INF = 1e9;
bool vis[MAXN];
int cost[MAXN][MAXN], mincost[MAXN], n;
int prim() {
    int ans = 0;memset(vis, false, sizeof(vis));
    for (int i = 0; i <= n; i++) mincost[i] = INF;
    mincost[0] = 0;
    while (true) {
        int v = -1;
        for (int u = 0; u < n; u++) {
            if (!vis[u] && (v == -1 || mincost[v] > mincost[u]))
                v = u;
        }
        if (v == -1) break;
        vis[v] = true; ans = max(mincost[v], ans);
        for (int u = 0; u < n; u++) {
            mincost[u] = min(mincost[u], cost[u][v]);
        }
    }
    return ans;
}
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                scanf("%d", &cost[i][j]);
                cost[j][i] = cost[i][j];
            }
        }
        printf("%d\n", prim());
    }
    return 0;
}

转载于:https://www.cnblogs.com/cniwoq/p/6770867.html