Hust oj 1101 Bombs of HRBUST(Prim)

最新推荐文章于 2017-05-04 12:29:00 发布

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Bombs of HRBUST

Time Limit: 1000 MS

Memory Limit: 65536 K

Total Submit: 184(91 users)

Total Accepted: 107(84 users)

Rating:

Special Judge: No

Description

Terrorists have placed bombs in each building of HRBUST during Summer Holiday.
They plan to assign fuse to make all the bombs connected so that they can destroy HRBUST just by detonating only one bomb, and the terrorists need 1 minute of time to assign 1 meter of fuse.
If bomb x and bomb y are connected by fuse, and one of them exploded, the other will explode immediately.(So they don't need to connect each pair of bombs.)
Now, you have got the distance between each pair of bombs, and you need to calculate the minimal time in minute they need to make all the bombs connected.

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases. Then T test cases follow.
For each test case:
Line 1. There are one number n(2≤n≤100), meaning the number of bombs.
Line 2..N + 1: There are one n * n matrix MAT (1≤MATij≤1000), the element MATij means the distance between bomb i and bomb j.
(It is guaranteed that the element on the diagonal line will always be 0 and MATij = MATji)

Output

For each test case:
Line 1. Output the minimal time in minute the terrorists need to achieve their goal.

Sample Input

2
2
0 1
1 0
3
0 2 1
2 0 3
1 3 0

Sample Output

1
3

裸Prim

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<queue>
using namespace std;

const int maxn = 305;
const int Inf = 0x3f3f3f;
int dis[maxn];
int map[maxn][maxn];
int vis[maxn];
int N;

void init()
{
    for(int i=0;i<N;i++)
    {
        dis[i] = map[i][0];
        vis[i] = 0;
    }
} //距离和访问数组初始化

int prim()
{
    init();
    dis[0] = 0; //起点距离初始化为0
    vis[0] = 1; //标记起点
    int ans = 0; //权值和初始化为0

    for(int i=0;i<N-1;i++)
    {
        int temp = Inf; // 记录最小边
        int flag; //纪录最小点
        for(int j=0;j<N;j++)
        {
            if(!vis[j] && dis[j] < temp)
            {
                temp = dis[j];
                flag = j;
            }
        } //找最小边
        vis[flag] = 1; //标记选出的边
        ans += dis[flag]; //把边加入生成树中
        for(int j=0;j<N;j++)
        {
            if(!vis[j] && dis[j] > map[flag][j])
                dis[j] = map[flag][j]; //更新最小权值
        }
    }
    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",&map[i][j]); //邻接矩阵存图
        }
        int ans = prim();
        printf("%d\n",ans);
    }
    return 0;
}