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最新推荐文章于 2019-03-18 23:55:47 发布

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#状压dp

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本文介绍了一种使用状态压缩动态规划（状压DP）的方法来解决草地种植问题，即在有限制条件下计算不同种植方案的数量。考虑到草地的大小限制，通过逐行计算可行方案并利用位运算进行状态转移，最终得出所有可能的种植方式。

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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

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.

第一道状压dp，看了别人的思路自己写出来了.题目大意是给你一块地，有能放牛的草地也有不能放牛的草地，而且两头牛不能同时安排在相邻的草地上，问有多少种放法，草地长和宽都小于等于12。

这题的思路是这样：先求出每一行的可以放的方案数，用num[i][j]记录下来各个方案所对应的十进制数，用num1[i]记录第i行的方案数，再初始化第一行dp[1][k]为1，然后使i从2到n循环，用dp[i][j]表示第i行第j种放置状态下的总方案数，依次累加，动规方程为dp[i][j]=dp[i][j]+(num[i][j]&num[i-1][k])?0:dp[i-1][k].(k=1,2,...num1[i-1])。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
int num[15][5000],n,m,num1[15],dp[15][5000];
void hang(int i,int temp)
{
	int t=0,j;
	for(j=0;j<(1<<m);j++){
		if(j&(j<<1))continue;
		if(j&temp)continue;
		t++;num[i][t]=j;
	}
	num1[i]=t;
}

int main()
{
	int i,j,c,temp,k,sum;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(num,0,sizeof(num));
		memset(num1,0,sizeof(num1));
		memset(dp,0,sizeof(dp));
		for(i=1;i<=n;i++){
			temp=0;
			for(j=1;j<=m;j++){
				scanf("%d",&c);
				c=1-c;
				temp=temp*2+c;
			}
			hang(i,temp);
		}
        for(j=1;j<=num1[1];j++){
        	dp[1][j]=1;
        }
        for(i=2;i<=n;i++){
        	for(j=1;j<=num1[i];j++){
	        	for(k=1;k<=num1[i-1];k++){
	        		if(num[i-1][k]&num[i][j])continue;
	        		dp[i][j]+=dp[i-1][k];
	        	}
	        }
        }
        sum=0;
        for(j=1;j<=num1[n];j++){
        	sum=(sum+dp[n][j])%100000000;
        }
        printf("%d\n",sum);
	}
	return 0;
}