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 34 ?

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.

第一道狀壓dp，看了別人的思路自己寫出來了.題目大意是給你一塊地，有能放牛的草地也有不能放牛的草地，而且兩頭牛不能同時安排在相鄰的草地上，問有多少種放法，草地長和寬都小于等于12。

這題的思路是這樣：先求出每一行的可以放的方案數，用num[i][j]記錄下來各個方案所對應的十進制數，用num1[i]記錄第i行的方案數，再初始化第一行dp[1][k]為1，然后使i從2到n循環，用dp[i][j]表示第i行第j種放置狀態下的總方案數，依次累加，動規方程為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; }

轉載于:https://www.cnblogs.com/herumw/p/9464739.html

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