HDU1054-Strategic Game
Strategic Game
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4114 Accepted Submission(s): 1824
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<vector>
#include<cstring>
#define N 1505
using namespace std;
int dp[N][2],visit[N];
vector<int> V[N];
int min(int a,int b)
{
return a>b?b:a;
}
void dfs(int k)
{
int i;
for(i=0;i<V[k].size();i++)
{
if(visit[V[k][i]] == 0) //判断是否为子节点
{
visit[ V[k][i] ] = 1;
dfs(V[k][i]);
visit[ V[k][i] ] = 0;
}
}
int s1=0,s2=0;
for(i=0;i<V[k].size();i++)
{
if(visit[V[k][i]] == 0) //判断是否为子节点
{
s1 += min(dp[ V[k][i] ][0] , dp[V[k][i]][1]);
s2 += dp[ V[k][i] ][1];
}
}
dp[k][1] = s1+1;
dp[k][0] = s2;
}
int main()
{
int n,i,j,t;
char a[20];
while(scanf("%d",&n)!=EOF)
{
memset(visit,0,sizeof(visit));
visit[1]=1;
for(i=0;i<n;i++)
V[i].clear();
for(i=0;i<n;i++)
{
scanf("%s",a);
int len=strlen(a);
int cc=0,tt=0;
j=0;
while(a[j]!='(')
{
if(a[j]>='0' && a[j]<='9')
cc=cc*10+a[j]-'0';
j++;
}
while(a[j]!=')')
{
if(a[j]>='0' && a[j]<='9')
tt=tt*10+a[j]-'0';
j++;
}
for(j=0;j<tt;j++)
{
scanf("%d",&t);
V[cc].push_back(t);
V[t].push_back(cc);
}
}
memset(dp,0,sizeof(dp));
dfs(1);
printf("%d\n",min(dp[1][0] , dp[1][1]));
}
return 0;
}
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