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

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 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:

 

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

 

 

第一道树形DP，感觉就是考察树的遍历，DP还是比较简单的

 

#include<stdio.h>
#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;
}