• HDU——T 1054 Strategic Game


    http://acm.hdu.edu.cn/showproblem.php?pid=1054

    Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 8510    Accepted Submission(s): 4096


    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   |   We have carefully selected several similar problems for you:  1053 1151 1281 1142 1233 
     
    矩阵跑不过去了、、
     1 #include <cstring>
     2 #include <cstdio>
     3 
     4 const int N(2555);
     5 int n,sumedge,head[N];
     6 struct Edge
     7 {
     8     int v,next;
     9     Edge(int v=0,int next=0):v(v),next(next){}
    10 }edge[233333];
    11 inline void ins(int u,int v)
    12 {
    13     edge[++sumedge]=Edge(v,head[u]);
    14     head[u]=sumedge;
    15     edge[++sumedge]=Edge(u,head[v]);
    16     head[v]=sumedge;
    17 }
    18 
    19 int match[N];
    20 bool vis[N];
    21 bool find(int u)
    22 {
    23     for(int v,i=head[u];i;i=edge[i].next)
    24         if(!vis[v=edge[i].v])
    25         {
    26             vis[v]=1;
    27             if(!match[v]||find(match[v]))
    28             {
    29                 match[v]=u;
    30                 return true;
    31             }
    32         }
    33     return false;
    34 }
    35 
    36 inline void read(int &x)
    37 {
    38     x=0; register char ch=getchar();
    39     for(;ch>'9'||ch<'0';) ch=getchar();
    40     for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
    41 }
    42 
    43 int main()
    44 {
    45     for(;~scanf("%d",&n);)
    46     {
    47         int ans=0;sumedge=0;
    48         for(int u,k,i=0;i<n;i++)
    49         {
    50             read(u),read(k);
    51             for(int v;k--;)
    52                 read(v),ins(u+1,v+1);
    53         }
    54         for(int i=1;i<=n;i++)
    55         {
    56             if(find(i)) ans++;
    57             memset(vis,0,sizeof(vis));
    58         }
    59         memset(edge,0,sizeof(edge));
    60         memset(head,0,sizeof(head));
    61         memset(match,0,sizeof(match));
    62         printf("%d
    ",ans>>1);
    63     }
    64     return 0;
    65 }
    ——每当你想要放弃的时候,就想想是为了什么才一路坚持到现在。
  • 相关阅读:
    前端面试
    (知乎)我想问一下PHP的学习路线图
    sublime text3如何安装bootstrap的插件
    php项目开发经验-2个月学习php经历
    基于链接的基本排序算法原理
    焦大:特征权重的处理与最终排名(中)
    焦大:特征权重的处理与最终排名(下)
    如何分析和监测竞争对手网站的seo数据
    没收录的页面是否会传递权重
    想学习SEO可以看哪些书籍
  • 原文地址:https://www.cnblogs.com/Shy-key/p/7435716.html
Copyright © 2020-2023  润新知