zoukankan      html  css  js  c++  java
  • HDU 1054 Strategic Game

    Strategic Game

    Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3580    Accepted Submission(s): 1573


    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
     
    Recommend
    JGShining

     1,二部图:建图要注意,否则会超时

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<vector>
    
    using namespace std;
    
    const int N=1510;
    
    vector<int> mp[N];
    int n,linker[N],vis[N];
    
    int DFS(int u){
        int v;
        for(v=0;v<(int)mp[u].size();v++)
            if(!vis[mp[u][v]]){
                vis[mp[u][v]]=1;
                if(linker[mp[u][v]]==-1 || DFS(linker[mp[u][v]])){
                    linker[mp[u][v]]=u;
                    return 1;
                }
            }
        return 0;
    }
    
    int Hungary(){
        int u,ans=0;
        memset(linker,-1,sizeof(linker));
        for(u=0;u<n;u++){
            memset(vis,0,sizeof(vis));
            if(DFS(u))
                ans++;
        }
        return ans;
    }
    
    int main(){
    
        //freopen("input.txt","r",stdin);
    
        while(~scanf("%d",&n)){
            int u,v,k;
            for(int i=0;i<n;i++)
                mp[i].clear();
            for(int i=0;i<n;i++){
                scanf("%d:(%d)",&u,&k);
                while(k--){
                    scanf("%d",&v);
                    mp[u].push_back(v);
                    mp[v].push_back(u);
                }
            }
            printf("%d\n",Hungary()/2);
        }
        return 0;
    }
  • 相关阅读:
    linux 进程
    VFS dup ,dup2
    文件操作 之 各类函数
    文件系统之 stat与access
    xml文件
    Java学习笔记42(数据库连接池 druid连接池)
    java学习笔记41(数据库连接池 C3p0连接池)
    java学习笔记39(sql事物)
    java学习笔记38(sql注入攻击及解决方法)
    java学习笔记37(sql工具类:JDBCUtils)
  • 原文地址:https://www.cnblogs.com/jackge/p/3090442.html
Copyright © 2011-2022 走看看