zoukankan      html  css  js  c++  java
  • hdu 1054 Strategic Game(tree dp)

    Strategic Game

    Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3806    Accepted Submission(s): 1672

    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
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    #define N 2010
    int Min(int i,int j){return i<j?i:j;}
    struct node{
    	node *l,*r;//l-son,r-brother
    	int f[2];
    	void init(){
    		l=r=NULL;f[0]=f[1]=0;
    	}
    }t[N],*rt;
    int r[N];
    void dfs(node *x){
    	//遍历x儿子
    	node *l=x->l;
    	while(l){
    		dfs(l);
    		//放在x上
    		x->f[1]+=Min(l->f[0],l->f[1]);
    		//x不放
    		x->f[0]+=l->f[1];
    		l=l->r;
    	}
    }
    int main(){
    	int i,j,k,x,y,n,m;
    	while(~scanf("%d",&n)){
    		m=0;
    		memset(r,-1,sizeof(r));
    		while(n--){
    			scanf("%d:(%d)",&i,&k);
    			if(r[i]==-1){r[i]=m;t[m++].init();}
    			x=r[i];
    			while(k--){//i->l=j
    				scanf("%d",&j);
    				if(r[j]==-1){r[j]=m;t[m++].init();}
    				y=r[j];
    				t[y].f[1]=1;
    				t[y].r=t[x].l;
    				t[x].l=&t[y];
    			}
    		}
    		for(i=0;;i++)
    			if(t[r[i]].f[1]==0){
    				rt=&t[r[i]];
    				rt->f[1]=1;
    				break;
    			}
    		dfs(rt);
    		printf("%d
    ",Min(rt->f[0],rt->f[1]));
    	}
    return 0;
    }

  • 相关阅读:
    linux2.6.24.3下移植SD/MMC到S3C2440上的全历程
    设置装备布置了下双表示器
    Virtual Box 1.5.0 - 实用的“无缝窗口”
    oracle DB LINK 运用
    Linux下的tidy安置
    Linux效劳器装机安全疾速进阶指南(2)
    Linux下历程间通信
    Firefox 3 CSS Hack
    Linux下的搜刮东西find根基用法
    Linux效能器装机平安快速进阶指南(3)
  • 原文地址:https://www.cnblogs.com/dyllove98/p/3238839.html
Copyright © 2011-2022 走看看