hdu1054 树形dp&&二分图

最新推荐文章于 2021-08-09 20:00:44 发布

转载最新推荐文章于 2021-08-09 20:00:44 发布 · 85 阅读

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原文链接：http://www.cnblogs.com/13224ACMer/p/5317167.html

本文介绍了一道关于在树形结构中部署最少数量的士兵以观察所有边的算法问题。采用树形动态规划的方法求解，确保任意路径都有士兵驻守。通过递归遍历节点并更新状态，实现最小化士兵数量的目标。

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B - Strategic Game

Time Limit:10000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1054

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:

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

      前几天刚做过这题，二分图A的，树形dp同样可以解决
     

      注意    此题要求任意一条路径上都应当有人看守，所有dp[root][0]+=dp[k][1],,,此处0的时候只能选着儿子节点的1
     

      dp[root][1]=min(dp[k][0],dp[k][1])
     

      题很水不多说了
     

 
       #include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1505;
struct node{
    int to,net;
}edge[maxn<<1];

int head[maxn];
bool  vis[maxn];
int dp[maxn][2];
int tot;

void addedge(int u,int v){

     edge[tot].to=v;
     edge[tot].net=head[u];
     head[u]=tot++;

     edge[tot].to=u;
     edge[tot].net=head[v];
     head[v]=tot++;

}

void dfs(int root){
    vis[root]=true;
    
    for(int i=head[root];i!=-1;i=edge[i].net){
        int v=edge[i].to;
        if(!vis[v]){
            dfs(v);
            dp[root][0]+=dp[v][1];
            dp[root][1]+=min(dp[v][1],dp[v][0]);

        }
    }
}

int main(){
    int t;
    while(scanf("%d",&t)!=EOF){
         memset(dp,0,sizeof(dp));
         memset(vis,false,sizeof(vis));
         tot=0;
         memset(head,-1,sizeof(head));
         for(int i=1;i<=t;i++){
                int n,x,y;
            scanf("%d:(%d)",&x,&n);
            for(int j=1;j<=n;j++){
                scanf("%d",&y);
                addedge(x,y);
            }
         }
         int root=0;
           for(int i=0;i<t;i++)
              dp[i][1]=1;
         dfs(root);
         int ans=min(dp[root][0],dp[root][1]);
        printf("%d\n",ans);
    }
    return 0;
} 
      
 

转载于:https://www.cnblogs.com/13224ACMer/p/5317167.html