zoukankan      html  css  js  c++  java
  • Corn Fields POJ

    Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares are infertile and can't be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two chosen squares share an edge. He has not yet made the final choice as to which squares to plant.

    Being a very open-minded man, Farmer John wants to consider all possible options for how to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways he can choose the squares to plant.

    Input

    Line 1: Two space-separated integers: M and N
    Lines 2.. M+1: Line i+1 describes row i of the pasture with N space-separated integers indicating whether a square is fertile (1 for fertile, 0 for infertile)

    Output

    Line 1: One integer: the number of ways that FJ can choose the squares modulo 100,000,000.

    Sample Input

    2 3
    1 1 1
    0 1 0

    Sample Output

    9

    Hint

    Number the squares as follows:
    1 2 3
      4  

    There are four ways to plant only on one squares (1, 2, 3, or 4), three ways to plant on two squares (13, 14, or 34), 1 way to plant on three squares (134), and one way to plant on no squares. 4+3+1+1=9.
    题解:dp[ i ][ state ]表示第i行当前状况下的方案数(状压dp的套路,只是需要一些二进制的技巧)
     1 #include<cstdio>
     2 #include<cstring>
     3 #include<iostream>
     4 #include<algorithm>
     5 using namespace std;
     6 
     7 const int maxn=13;
     8 const int mod=1e8;
     9 
    10 int n,m;
    11 int dp[maxn][1<<maxn],line[1<<maxn];
    12 
    13 int main()
    14 {   cin>>n>>m;
    15     for(int i=1;i<=n;i++){
    16         for(int j=1;j<=m;j++){
    17             int temp;
    18             cin>>temp;
    19             line[i]|=temp<<(j-1);
    20         } 
    21     }
    22     
    23     dp[0][0]=1;
    24     for(int i=1;i<=n;i++){
    25         for(int j=0;j<(1<<m);j++){
    26             if((j&line[i])==j&&(j&(j>>1))==0){
    27                 for(int k=0;k<(1<<m);k++) if((j&k)==0) dp[i][j]=(dp[i][j]+dp[i-1][k])%mod; 
    28             }
    29         }
    30     }
    31     
    32     int ans=0;
    33     for(int i=0;i<(1<<m);i++) ans=(ans+dp[n][i])%mod;
    34     cout<<ans<<endl;
    35 }
  • 相关阅读:
    阿里云 oss (二) 生成令牌
    doc转html
    js 实现下载本地文件
    vue 中使用video 使用视频/嵌入视频
    小程序裁剪
    车牌号正则
    小程序单选框 radio
    小程序省市三级联动 及日期选择 (年月日)
    小程序图片选择,小程序图片上传及调用后台接口上传
    element ui省市区三级联动,及获取三级联动的name值
  • 原文地址:https://www.cnblogs.com/zgglj-com/p/7806619.html
Copyright © 2011-2022 走看看