zoukankan      html  css  js  c++  java
  • 【POJ3254】Corn Fields 状压DP第一次

    !!!!!!!

    第一次学状压DP,其实就是运用位运算来实现一些比较,挺神奇的。。

    为什么要发“!!!”因为!x&y和!(x&y)。。感受一下。。

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #define N 13
     5 #define M 1<<13
     6 #define MOD 1000000000
     7 using namespace std;
     8 int n,m,t,ans;
     9 int so[M],map[M];
    10 int dp[N][M];
    11 bool ju1(int x) {return x&(x<<1);}
    12 bool ju2(int x,int y) {return map[x]&so[y];}
    13 int main()
    14 {
    15     while (~scanf("%d%d",&n,&m))
    16     {
    17     memset(map,0,sizeof(map));
    18     memset(so,0,sizeof(so));
    19     memset(dp,0,sizeof(dp));
    20     ans=0; 
    21     t=0;
    22     for (int i=1;i<=n;i++)
    23         for (int j=1;j<=m;j++)
    24         {
    25             int x;
    26             scanf("%d",&x);
    27             if (x==0) map[i]+=(1<<(j-1));
    28         }
    29     for (int i=0;i<(1<<m);i++)
    30         if (!ju1(i))    so[t++]=i;
    31     for (int i=0;i<t;i++)
    32         if (!ju2(1,i)) dp[1][i]=1;
    33     for (int i=2;i<=n;i++)
    34         for (int j=0;j<t;j++)
    35         {
    36             if (ju2(i,j)) continue;
    37             for (int k=0;k<t;k++)
    38             {
    39                 if (ju2(i-1,k)) continue;
    40                 if (!(so[j]&so[k])) dp[i][j]+=dp[i-1][k];
    41             }
    42         }
    43     //for (int i=1;i<=n;i++)
    44     //printf("%d
    ",t);
    45     for (int j=0;j<t;j++)
    46             ans=(ans+dp[n][j])%MOD;
    47     printf("%d
    ",ans);
    48     }
    49     return 0;
    50 }
    View Code

    Description

    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.

    Source

    —Anime Otaku Save The World.
  • 相关阅读:
    centos8 防火墙配置增加端口
    linux上搭建maven私服(下)
    linux上搭建maven私服(中)
    项目成本管理中的PV、EV与AC的区别-实例解释
    配置IKE SA的生存周期(华为)
    IKE SA和IPSec SA的区别
    IPsecVPN协商过程-主模式
    Fortigate防火墙常用命令
    飞塔防火墙清除系统密码
    fatal: unable to access ‘https://github xxxxxxxxx的解决方法
  • 原文地址:https://www.cnblogs.com/DMoon/p/5335014.html
Copyright © 2011-2022 走看看