Strategic Game
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6421 Accepted Submission(s): 2987
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 description of each node in the following format
node_identifier:(number_of_roads) node_identifier1 node_identifier2 ... node_identifier
or
node_identifier:(0)
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:
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)
2
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#define mem(x,y) memset(x,y,sizeof(x))
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=;
vector<int>vec[MAXN];
int vis[MAXN],usd[MAXN];
bool dfs(int u){
for(int i=;i<vec[u].size();i++){
int v=vec[u][i];
if(!vis[v]){
vis[v]=;
if(usd[v]==-||dfs(usd[v])){
usd[v]=u;return true;
}
}
}
return false;
}
int main(){
int N;
while(~scanf("%d",&N)){
for(int i=;i<N;i++)vec[i].clear();
int a,t,b;
for(int i=;i<N;i++){
scanf("%d:(%d)",&a,&t);
while(t--){
scanf("%d",&b);vec[a].push_back(b);
vec[b].push_back(a);
}
}
mem(vis,);mem(usd,-);
int ans=;
for(int i=;i<N;i++){
mem(vis,);
if(dfs(i))ans++;
}
printf("%d\n",ans/);
}
return ;
}