1004 Strategic game 最小点覆盖

 链接：https://ac.nowcoder.com/acm/contest/25022/1004
来源：牛客网

题目描述

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. 

For example for the tree: 

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

输入描述:

The input 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_identifier1node\_identifier_1node_identifier1​ node_identifier2node\_identifier_2node_identifier2​ ... node_identifiernumberofroadsnode\_identifier_{number_of_roads }node_identifiernumbero​fr​oads​
or 
node_identifier:(0) 
The node identifiers are integer numbers between 0 and n-1, for n nodes (0<n≤1500)(0 \lt  n \leq 1500)(0<n≤1500);the number_of_roads in each line of input will no more than 10. Every edge appears only once in the input data.

输出描述:

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:

示例1

输入

复制 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)

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)

输出

复制 1 2

1
2

分析

树的最小点覆盖：每两个相连的结点之间至少有一个节点被覆盖。

f[i][1]表示选择这个点，f[i][0] 表示不选择这个点 

状态转移：f[i][1] = 1 + E min(f[k][0],f[k][1]);k表示子节点

f[i][0] = E f[k][1];

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
const int N = 1510;
int h[N],e[N],w[N],ne[N],idx;
int n;
int f[N][N];
bool st[N];
void add(int a,int b)
{
    e[idx] = b,ne[idx] = h[a],h[a] = idx++;
}

void dfs(int u)
{
    f[u][0] = 0;
    f[u][1] = 1;
    for(int i = h[u]; i != -1; i = ne[i])
    {
        int j = e[i];
        dfs(j);
        f[u][0] += f[j][1];
        f[u][1] += min(f[j][0],f[j][1]);
    }
    
}


int main()
{
    // ios::sync_with_stdio(false);
    // cin.tie(nullptr);
    while(cin >> n)
    {
        memset(h, -1, sizeof h);
        memset(st, 0, sizeof st);
        idx = 0;
        for(int i = 0; i < n; i ++ )
        {
            int id,cnt;
            scanf("%d:(%d)",&id,&cnt);
            while(cnt -- )
            {
                int ver;
                cin >> ver;
                add(id, ver);
                st[ver] = true;
            }
        }
        int r = 0;
        while(st[r]) r++;
        dfs(r);
        cout << min(f[r][0],f[r][1]) << "\n";
    }
    return 0;   
}