Strategic Game |
| Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 141 Accepted Submission(s): 96 |
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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 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: |
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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) |
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Sample Output 1 2 |
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Source Southeastern Europe 2000 |
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Recommend JGShining |
/*
初步思想：树状DP+搜索
加一个标记现在这个节点放不放士兵
1A,哈哈哈哈
*/
#include<bits/stdc++.h>
#define N 1505
using namespace std;
vector<int>edge[N];
int n;
int u,m,v;
int dp[N][2];//dp[i][1/0]表示以i为根节点放士兵和不放士兵的最小值
void dfs(int root,int fa){
dp[root][1]=1;
dp[root][0]=0;
for(int i=0;i<edge[root].size();i++){
int v=edge[root][i];
if(v==fa) continue;
dfs(v,root);
dp[root][0]+=dp[v][1];
dp[root][1]+=min(dp[v][0],dp[v][1]);
}
}
int main(){
//freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
while(scanf("%d",&n)!=EOF){
memset(dp,0,sizeof dp);
for(int i=0;i<=n;i++)
edge[i].clear();
for(int i=0;i<n;i++){
scanf("%d:(%d)",&u,&m);
for(int j=0;j<m;j++){
scanf("%d",&v);
edge[u].push_back(v);
edge[v].push_back(u);
}
}//建图
// for(int i=0;i<n;i++){
// cout<<edge[i].size()<<endl;
// }
dfs(0,-1);
printf("%d\n",min(dp[0][0],dp[0][1]));
}
return 0;
}