UVALive 2038 Strategic game （树形DP，4级）_strategic game uvalive

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L - Strategic game

Crawling in process... Crawling failed Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

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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.

For example for the tree:

the solution is one soldier (at the node 1).

Input

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_identifiernumber_of_roads
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.

Output

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).

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

思路：简单树形DP dp[u][0]不选dp[u][1]选。

     dp[u][1]+=min(dp[v][0],dp[v][1])

     dp[u][0]+=dp[v][1];

数据生成器

#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cstdlib>
using namespace std;
bool vis[1600][1600];
int main()
{
  int n;
  memset(vis,0,sizeof(vis));
  for(int i=0;i<=1503;++i)vis[i][i]=1;
  srand((int)time(0));
  n=rand()%1500+1;printf("%d\n",n);
  int z=n;
  for(int i=0;i<n;++i)
  {
    int zz=rand()%100;
    if(z>zz)
    z-=zz;
    else zz=0;
    printf("%d:(%d)",i,zz);
    int k=0;
    for(int j=0;j<zz;++j)
    { while(k<n&&vis[i][k])++k;
      printf(" %d",k);vis[i][k]=vis[k][i]=1;
    }
     printf("\n");
  }
}

对拍程序

@echo off
:loop
 rand.exe>data.txt
 std.exe<data.txt>std.txt
 my.exe<data.txt>my.txt
 fc my.txt std.txt
 if not errorlevel 1 goto loop
 pause 
 goto loop

代码

#include<cstring>
#include<iostream>
#include<cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=1502;
const int oo=0x3f3f3f3f;
int head[mm],edge;
int dp[mm][2];
class Edge
{
  public:int v,next;
}e[mm*mm];
void data()
{
  edge=0;clr(head,-1);
}
void add(int u,int v)
{
  e[edge].v=v;e[edge].next=head[u];head[u]=edge++;
}
int DP(int u,int fa,int yes)
{ int v;
  if(dp[u][yes]!=-1)return dp[u][yes];
  dp[u][yes]=yes;
  for(int i=head[u];~i;i=e[i].next)
  {
    v=e[i].v;
    if(v==fa)continue;
    if(yes)dp[u][yes]+=min(DP(v,u,0),DP(v,u,1));
    else dp[u][yes]+=DP(v,u,1);
  }
  return dp[u][yes];
}
int main()
{
  int n,a,b,m;

  while(~scanf("%d",&n))
  {
    data();
    FOR(i,1,n)
    {
      scanf("%d",&a);
      scanf(":(%d)",&m);
      FOR(j,1,m)
      {
        scanf("%d",&b);
        add(a,b);add(b,a);
        //cout<<a<<" "<<b<<endl;
      }
    }
    clr(dp,-1);
    int ans=min(DP(0,-1,0),DP(0,-1,1));
    printf("%d\n",ans);
  }
}