poj 1463 Strategic game DP

题目地址：http://poj.org/problem?id=1463

题目：

Strategic game

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 7929		Accepted: 3692

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. 

For example for the tree: 

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

Input

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_identifier₁ node_identifier₂ ... node_identifier_{number_of_roads} 
  or 
  node_identifier:(0) 

The node identifiers are integer numbers between 0 and n-1, for n nodes (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.

Output

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:

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

Source

Southeastern Europe 2000

 

思路：明显树形dp嘛，然而写完一交，MLE的我不省人事，这题目居然这么变态卡内存，，

　　　我把记录边的vector改成链式前向星就ac了

　　　这个和上题思路一样，就不啰嗦了，可以去看看前一篇树形dp的

代码：　

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>

#define PB push_back
#define MP make_pair
using namespace std;
typedef long long LL;
#define PI acos((double)-1)
#define E exp(double(1))
const int K=1500+2;
short vis[K],head[K],cnt;
struct Edge
{
    short next,to;
}edge[K*2];
void add(int u,int v)
{
    edge[cnt].next=head[u];
    edge[cnt].to=v;
    head[u]=cnt++;
}
void dfs(int x,int &v1,int &v2)
{
    v1=1;v2=0;
    vis[x]=1;
    int x1,x2;
    for(int i=head[x];~i;i=edge[i].next)
    {
        int v=edge[i].to;
        if(vis[v])continue;
        dfs(v,x1,x2);
        v1+=min(x1,x2);
        v2+=x1;
    }
}
int main(void)
{
    int n;
    while(~scanf("%d",&n))
    {
       cnt=0;
       memset(head,-1,sizeof(head));
       memset(vis,0,sizeof(vis));
       for(int i=1;i<=n;i++)
       {
           int u,v,k;
           scanf("%d:(%d)",&u,&k);
           while(k--)
            scanf("%d",&v),add(u,v),add(v,u);
       }
       int x,y;
       dfs(0,x,y);
       printf("%d\n",min(x,y));
    }
    return 0;
}