hdu 1054 Strategic Game（最小顶点覆盖）

最新推荐文章于 2025-09-18 21:14:24 发布

最新推荐文章于 2025-09-18 21:14:24 发布 · 311 阅读

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本文探讨了最小顶点覆盖问题的解决方法，通过构建树形结构并使用邻接表来实现高效的算法求解。重点介绍了算法的核心思想、输入输出格式及样例解析。

Strategic Game

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1760 Accepted Submission(s): 742

Problem Description

Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and then he is very sad. Now he has the following problem. He must defend a medieval city, the roads of which form a tree. He has to put the minimum number of soldiers on the nodes so that they can observe all the edges. Can you help him?

Your program should find the minimum number of soldiers that Bob has to put for a given tree.

The input file contains several data sets in text format. Each data set represents a tree with the following description:

the number of nodes
the description of each node in the following format
node_identifier:(number_of_roads) node_identifier1 node_identifier2 ... node_identifier
or
node_identifier:(0)

The node identifiers are integer numbers between 0 and n-1, for n nodes (0 < n <= 1500). Every edge appears only once in the input data.

For example for the tree:

the solution is one soldier ( at the node 1).

The output should be printed on the standard output. For each given input data set, print one integer number in a single line that gives the result (the minimum number of soldiers). An example is given in the following table:

Sample Input

  4 0:(1) 1 1:(2) 2 3 2:(0) 3:(0) 5 3:(3) 1 4 2 1:(1) 0 2:(0) 0:(0) 4:(0)
 

Sample Output

1 2

典型的最小顶点覆盖！最小顶点覆盖 == 最大匹配（双向图）/2；

此题有个小细节，数据较大，要用邻接表，不然会超时！

连接：http://acm.hdu.edu.cn/showproblem.php?pid=1054

代码：

#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <vector>
using namespace std;

const int N = 1505;

int pre[N];
bool flag[N];
vector<int> map[N];
int n;

int find(int cur)
{
    int i, k;
    for(i = 0; i < map[cur].size(); i++)
    {
        k = map[cur][i];
        if(!flag[k])
        {
            flag[k] = true;
            if(pre[k] == -1 || find(pre[k]))
            {
                pre[k] = cur;
                return 1;
            }
        }
    }
    return 0;
}

int main()
{
    int i, j, r, k, num, sum;
    while(scanf("%d", &n) != EOF)
    {
        memset(pre, -1, sizeof(pre));
        for(i = 0; i < n; i++) map[i].clear();
        for(i = 0; i < n; i++)
        {
            scanf("%d:(%d)", &k, &num);
            for(j = 0; j < num; j++)
            {
                scanf("%d", &r);
                map[k].push_back(r);	//用邻接表
                map[r].push_back(k);	//建双向图
            }
        }
        sum = 0;
        for(i = 0; i < n; i++)
        {
            memset(flag, false, sizeof(flag));
            sum += find(i);
        }
        printf("%d\n", sum/2);
    }

    return 0;
}