zoukankan      html  css  js  c++  java
  • POJ1947 Rebuilding Roads

     
    Time Limit: 1000MS   Memory Limit: 30000K
    Total Submissions: 11334   Accepted: 5222

    Description

    The cows have reconstructed Farmer John's farm, with its N barns (1 <= N <= 150, number 1..N) after the terrible earthquake last May. The cows didn't have time to rebuild any extra roads, so now there is exactly one way to get from any given barn to any other barn. Thus, the farm transportation system can be represented as a tree. 

    Farmer John wants to know how much damage another earthquake could do. He wants to know the minimum number of roads whose destruction would isolate a subtree of exactly P (1 <= P <= N) barns from the rest of the barns.

    Input

    * Line 1: Two integers, N and P 

    * Lines 2..N: N-1 lines, each with two integers I and J. Node I is node J's parent in the tree of roads. 

    Output

    A single line containing the integer that is the minimum number of roads that need to be destroyed for a subtree of P nodes to be isolated. 

    Sample Input

    11 6
    1 2
    1 3
    1 4
    1 5
    2 6
    2 7
    2 8
    4 9
    4 10
    4 11
    

    Sample Output

    2

    Hint

    [A subtree with nodes (1, 2, 3, 6, 7, 8) will become isolated if roads 1-4 and 1-5 are destroyed.] 

    Source

     

    给你一棵有n个节点的树,可以砍掉其中一些边,问最少砍几条边可以留下一个有p个节点的树(剩下的必须是一整棵树)

    标准树形DP

     
    各种改都改不对,无奈开启了代码标准比对模式,仍然没用。
    这时我注意到不管怎么改输出的都是一样的解,md卡程序了!关了IDE,删了编译好的程序,开IDE重编译,输出了正解……
    然而我恐怕再也无法知道我到底是在哪一步改对了的。
     
     1 /*by SilverN*/
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<cmath>
     7 #include<vector>
     8 using namespace std;
     9 const int mxn=160;
    10 int read(){
    11     int x=0,f=1;char ch=getchar();
    12     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    13     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    14     return x*f;
    15 }
    16 int n,p;
    17 int f[mxn][mxn];
    18 int num[mxn];
    19 vector<int>e[mxn];
    20 void dp(int u){
    21     int i,j,k;
    22     num[u]=1;
    23     if(!e[u].size()){f[u][1]=0;return;}
    24     for(i=0;i<e[u].size();i++){
    25         int v=e[u][i];
    26         dp(v);
    27         num[u]+=num[v];
    28         for(j=num[u];j;j--){
    29             for(k=1;k<j;k++){
    30                 f[u][j]=min(f[u][j],f[u][j-k]+f[v][k]-1);
    31             }
    32         }
    33     }
    34     return;
    35 }
    36 int main(){
    37     n=read();p=read();
    38     int i,j;
    39     int u,v;
    40     for(i=1;i<n;i++){
    41         u=read();v=read();
    42         e[u].push_back(v);
    43         num[u]++;
    44     }
    45     memset(f,0x3f,sizeof f);
    46     for(i=1;i<=n;i++){
    47         f[i][1]=num[i];
    48     }
    49     dp(1);
    50     int ans=f[1][p];
    51     for(i=1;i<=n;i++){
    52         ans=min(ans,f[i][p]+1);
    53     }
    54     printf("%d
    ",ans);
    55     return 0;
    56 }
  • 相关阅读:
    数据结构 图的接口定义与实现分析(超详细注解)
    数据结构-优先队列 接口定义与实现分析
    什么是梯度?
    javascript当中火狐的firebug如何单步调试程序?
    给出一个javascript的Helloworld例子
    lvs dr 模型配置详解
    数据库连接池
    JDBC事务管理
    JDBC工具类:JDBCUtils
    详解JDBC对象
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/5918273.html
Copyright © 2011-2022 走看看