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

dp[子樹的根][是否擺放]
dp[][0] 不擺放, 則 dp[][0] += dp[所有子節點][1]
dp[][1] 擺放, 則 dp[][1] += dp[所有子節點的]min([0], [1]);
最後 dp[][1] += 1;

#include <stdio.h>
#include <vector>
#define min(x, y) ((x) < (y) ? (x) : (y))
using namespace std;
struct Arc {
    int to;
};
vector<Arc> nd[2000];
int dp[2000][2], used[2000];
int dfs(int node) {
    used[node] = 1;
    dp[node][0] = 0;
    dp[node][1] = 1;
    for(vector<Arc>::iterator i = nd[node].begin(); i != nd[node].end(); i++) {
        if(used[i->to] == 0) {
            dfs(i->to);
            dp[node][0] += dp[i->to][1];
            dp[node][1] += min(dp[i->to][0], dp[i->to][1]);
        }
    }
    return min(dp[node][0], dp[node][1]);
}
int main() {
    int n, m, st, ed, i;
    while(scanf("%d", &n) == 1) {
        for(i = 0; i < n; i++)
            nd[i].clear(), used[i] = 0;
        Arc tmp;
        for(i = 0; i < n; i++) {
            scanf("%d: (%d)", &st, &m);
            while(m--) {
                scanf("%d", &ed);
                tmp.to = ed;
                nd[st].push_back(tmp);
                tmp.to = st;
                nd[ed].push_back(tmp);
            }
        }
        printf("%d\n", dfs(0));
    }
    return 0;
}