hdu1054 树状dp

B - 树形dp

Crawling in process... Crawling failed Time Limit:2000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

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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 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_{number_of_roads}
  or
  node_identifier:(0)

The node identifiers are integer numbers between 0 and n-1, for n nodes (0 < n <= 1500);the number_of_roads in each line of input will no more than 10. 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). An example is given in the following:

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
题目大意：给你一个树状结构，每一个节点都可以放置士兵，可以看守住与 该节点相邻的所有边，问你最少需要多少个节点。
思路分析：树的最小点覆盖问题，首先确定状态f[i[0]表示在根节点i上不放置士兵看守住这棵子树需要的最少士兵，f[i][0]
则代表根结点i上不放置士兵看守住所需要的最少士兵，状态转移方程f[i][1]+=min(f[j][1],f[j][0]),f[i][0]+=f[j][1]
初始化f[i][[1]=1,f[i][0]=0;

代码：

/*树的最小点覆盖
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1500+10;
int fat[maxn];
int dp[maxn][2];
bool vis[maxn];//标记
int n;
void dfs(int x)
{
    //cout<<x<<endl;
    vis[x]=true;
    dp[x][1]=1;
    dp[x][0]=0;
    for(int i=0;i<n;i++)
    {
        if(!vis[i]&&fat[i]==x)
        {
            dfs(i);
            dp[x][1]+=min(dp[i][0],dp[i][1]);//子节点可放可不放
            dp[x][0]+=dp[i][1];
        }
    }
}
int main()
{
    int r,num;
    int a;
    int i;
    while(scanf("%d",&n)!=EOF)
    {
       // cout<<n<<endl;
        memset(fat,0,sizeof(fat));
        memset(vis,false,sizeof(vis));
        for(int k=0;k<n;k++)
        {
        scanf("%d:(%d)",&r,&num);
        //cout<<r<<" "<<num<<endl;
        for(int i=1;i<=num;i++)
        {
            scanf("%d",&a);
            fat[a]=r;
          // cout<<a<<endl;
        }
        }
       // for(int i=0;i<n;i++)
            //cout<<fat[i]<<endl;
        for(i=0;i<n;i++)
        {
            if(!fat[i])
            {
                //cout<<i<<endl;
                dfs(i);
                cout<<min(dp[i][0],dp[i][1])<<endl;
                break;
            }
        }

    }
    return 0;
}

　　思考了下我这样写是有弊端的，这是一个无向图，然而我建树的时候默认了先出的数为根，后出的为子节点，这样是不对的。。。然而数据水过，自己又用链式前向星

存图写了一些，效率大大提高

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=+;
struct node
{
    int to;
    int next;
};
node edge[*maxn];
int head[maxn];
int dp[maxn][];
bool vis[maxn];//把一个无向 图变成有向图
int tot;
int n;
void add(int x,int y)
{
    edge[tot].to=y;
    edge[tot].next=head[x];
    head[x]=tot++;
}
void init()
{
    memset(head,-,sizeof(head));
    memset(vis,false,sizeof(vis));
    tot=;
    int r,num;
    int a;
   for(int i=;i<n;i++)
   {
       scanf("%d:(%d)",&r,&num);
       for(int i=;i<num;i++)
       {
           scanf("%d",&a);
           add(a,r);
           add(r,a);
       }
   }
}
void dfs(int nod)
{
    vis[nod]=true;
    dp[nod][]=,dp[nod][]=;
    for(int i=head[nod];i!=-;i=edge[i].next)
    {
        int v=edge[i].to;//这条边指向的点
        if(!vis[v])
        {
            dfs(v);
            dp[nod][]+=min(dp[v][],dp[v][]);
            dp[nod][]+=dp[v][];
        }
    }
}
int main()
{
   while(scanf("%d",&n)!=EOF)
   {
       init();
       dfs();
       printf("%d\n",min(dp[][],dp[][]));
   }
}