poj 2485 Highways（最小產生樹）

標籤：poj   最小產生樹   

題目連結：http://poj.org/problem?id=2485

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

130 990 692990 0 179692 179 0

Sample Output

692

Hint

Huge input,scanf is recommended.

Source

POJ Contest,Author:[email protected]

題意：

尋找最小產生樹圖中最長的路徑！

和POJ 1258一樣的，加個判斷條件就可以了！

但是注意這題用memset 初始化visited會WA， 用for迴圈就不會了！POJ 1258 則不存在這個問題！

代碼如下：

#include <cstdio>#include <cstring>#define INF 0x3f3f3f3f#define MAXN 517//建立m二維數組儲存圖表，low數組記錄每2個點間最小權值，visited數組標記某點是否已訪問int m[MAXN][MAXN], low[MAXN], visited[MAXN];int n;int prim( ){    int i, j;    int maxx = 0;    int pos, minn, result=0;    // memset(visited,0,sizeof(visited));    for(i = 1; i <= n; i++)        visited[i] = 0;    visited[1] = 1;    pos = 1;          //從某點開始，分別標記和記錄該點    for(i = 1; i <= n; i++)     //第一次給low數組賦值    {        if(i != pos)            low[i] = m[pos][i];        else            low[i] = 0;    }    for(i = 1; i < n; i++) //再運行n-1次    {        minn = INF;   //找出最小權值並記錄位置        for(j = 1; j <= n; j++)        {            if(visited[j]==0 && minn>low[j])            {                minn = low[j];                pos = j;            }        }        // result += minn;   //最小權值累加        if(minn > maxx)            maxx = minn;        visited[pos] = 1;   //標記該點        for(j = 1; j <= n; j++)   //更新權值            if(visited[j]==0 && low[j]>m[pos][j])                low[j] = m[pos][j];    }    //return result;    return maxx;}int main(){    int i,j,ans;    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        memset(m,INF,sizeof(m));   //所有權值初始化為最大        for(i = 1; i <= n; i++)        {            for(j = 1; j <= n; j++)            {                scanf("%d",&m[i][j]);            }        }        ans = prim( );        printf("%d\n",ans);    }    return 0;}