HDU 1054.Strategic Game【动态规划、DP】【4月7】

Strategic Game

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6801    Accepted Submission(s): 3190

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

 

当前节点不放，则子节点一定放；当前节点放，子节点放或不放两种选择，选择优者。具体看代码：

#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
#define MAXN 1510
vector <int> NEXT[MAXN];
int N, t, fa, nxt, ans;
int s[MAXN], dp[MAXN][2];
int DFS(int node, int flag)
{
    if(dp[node][flag] != -1) return dp[node][flag];//用dp数组保存，减少重复运算次数
    int sum = flag;
    vector <int>::iterator it = NEXT[node].begin();
    for(;it != NEXT[node].end(); ++it)
    {
        if(flag == 0) sum += DFS(*it, 1);//当前不放，子节点一定得放
        else sum += min(DFS(*it, 0),DFS(*it, 1));//当前放，子节点有两种选择，取小值
    }
    dp[node][flag] = sum;//用dp数组保存，减少重复运算次数
    return sum;
}
int main()
{
    while(scanf("%d", &N) != EOF && N)
    {
        for(int i = 0;i < N; ++i)
        {
            dp[i][0] = dp[i][1] = -1;
            s[i] = i;
            NEXT[i].clear();
        }
        for(int i = 0;i < N; ++i)
        {
            scanf("%d:(%d)", &fa, &t);
            while(t--)
            {
                scanf("%d", &nxt);
                s[nxt] = fa;//保存父子关系，以便查找根节点
                NEXT[fa].push_back(nxt);
            }
        }
        for(int i = 0;i < N; ++i)
        {
            if(s[i] == i)//根节点
            {
                ans = min(DFS(i, 0), DFS(i, 1));
                break;
            }
        }
        cout << ans << endl;
    }
    return 0;
}