Uva 1292 - Strategic game 树形dp 最小点覆盖

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).

Input 

• the number of nodes
• the description of each node in the following format:
  

  node_identifier:(number_of_roads) node_identifier1 node_identifier2 � node_identifiernumber_of_roads
  or
  node_identifier:(0)

Output 

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

f[i][0]表示点i属于点覆盖，并且以点i为根的子树中所连接的边都被覆盖的情况下点覆盖集中所包含的最少点数。

f[i][1]表示点i不属于点覆盖，且以i为根的子树中所连接的边都被覆盖的情况下点覆盖集中所包含最少点的个数。

① f[i][0]=1+sum{ min(f[v][0], f[v][1]) };

② f[i][1]=sum{ f[v][0] };

-------------

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

const int maxn=3111;
const int INF=1e9;

struct EDGENODE
{
    int to,next;
} edges[maxn];
int head[maxn],edge;
void addedge(int u,int v)
{
    edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
    edges[edge].to=u,edges[edge].next=head[v],head[v]=edge++;
}
void init_edges()
{
    memset(head,-1,sizeof(head));
    edge=0;
}
int n;
int f[maxn][2];
void dp(int u,int pa)
{
    f[u][0]=1;
    f[u][1]=0;
    for (int k=head[u]; k!=-1; k=edges[k].next)
    {
        int v=edges[k].to;
        if (v==pa) continue;
        dp(v,u);
        f[u][0]+=min(f[v][0],f[v][1]);
        f[u][1]+=f[v][0];
    }
}

int main()
{
    while (~scanf("%d",&n))
    {
        int u,v,t;
        init_edges();
        for (int i=0; i<n; i++)
        {
            scanf("%d:(%d)",&u,&t);
            while (t--)
            {
                scanf("%d",&v);
                addedge(u,v);
            }
        }
        dp(0,-1);
        printf("%dn",min(f[0][0],f[0][1]));
    }
    return 0;
}

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