hdu 4804 Campus Design 插頭dp

Campus Design

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 515    Accepted Submission(s): 253

Problem Description

Nanjing University of Science and Technology is celebrating its 60th anniversary. In order to make room for student activities, to make the university a more pleasant place for learning, and to beautify the campus, the college administrator decided to start construction on an open space.
The designers measured the open space and come to a conclusion that the open space is a rectangle with a length of n meters and a width of m meters. Then they split the open space into n x m squares. To make it more beautiful, the designer decides to cover the open space with 1 x 1 bricks and 1 x 2 bricks, according to the following rules:

1. All the bricks can be placed horizontally or vertically
2. The vertexes of the bricks should be placed on integer lattice points
3. The number of 1 x 1 bricks shouldn’t be less than C or more than D. The number of 1 x 2 bricks is unlimited.
4. Some squares have a flowerbed on it, so it should not be covered by any brick. (We use 0 to represent a square with flowerbet and 1 to represent other squares)

Now the designers want to know how many ways are there to cover the open space, meeting the above requirements.

 

Input

There are several test cases, please process till EOF.
Each test case starts with a line containing four integers N(1 <= N <= 100), M(1 <= M <= 10), C, D(1 <= C <= D <= 20). Then following N lines, each being a string with the length of M. The string consists of ‘0’ and ‘1’ only, where ‘0’ means the square should not be covered by any brick, and ‘1’ otherwise.

 

Output

Please print one line per test case. Each line should contain an integers representing the answer to the problem (mod 10⁹ + 7).

 

Sample Input

1 1 0 0 1 1 1 1 2 0 1 1 1 2 1 1 2 1 2 11 1 2 0 2 01 1 2 0 2 11 2 2 0 0 10 10 2 2 0 0 01 10 2 2 0 0 11 11 4 5 3 5 11111 11011 10101 11111

 

Sample Output

0 0 1 1 1 2 1 0 2 954

 

Source

2013ACM/ICPC亞洲區南京站現場賽——題目重現

 

Recommend

liuyiding

 

題意：有1*1和1*2兩種磚，0點不能鋪磚，求用1*1磚數量在c到d的鋪磚種類數。

思路：加一維表示用1*1磚的數量，注意特判m，n=1的情況。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=1<<12;
const int mod=1e9+7;
int dp[2][N][25];
char vis[105][13];
int n,m,c,d;
int main()
{
    while(scanf("%d%d%d%d",&n,&m,&c,&d)!=EOF){
        for(int i=0;i<n;i++){
            getchar();
            scanf("%s",vis[i]);
        }
        if(n==1&&m==1){
            if(vis[0][0]=='1'&&c<=1&&d>=1)
             {
                printf("1\n");
                continue;
             }
            else {
                printf("0\n");
                continue;
            }
        }
        memset(dp,0,sizeof(dp));
        dp[0][0][0]=1;
        int pre=1,now=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                pre=!pre;
                now=!now;
                int q=1<<j;
                int p=q<<1;
                memset(dp[now],0,sizeof(dp[now]));
                for(int k=0;k<1<<(m+1);k++){
                    for(int l=0;l<=d;l++){
                        if(vis[i][j]=='1'){
                            if((k&p)&&!(k&q))
                                dp[now][k-p][l]=(dp[now][k-p][l]+dp[pre][k][l])%mod;
                            if(!(k&p)&&(k&q))
                                dp[now][k-q][l]=(dp[now][k-q][l]+dp[pre][k][l])%mod;
                            if(!(k&p)&&!(k&q)){
                                dp[now][k][l+1]=(dp[now][k][l+1]+dp[pre][k][l])%mod;
                                dp[now][k+p][l]=(dp[now][k+p][l]+dp[pre][k][l])%mod;
                                dp[now][k+q][l]=(dp[now][k+q][l]+dp[pre][k][l])%mod;
                            }
                        }
                        else if(!(k&p)&&!(k&q))
                            dp[now][k][l]=dp[pre][k][l];
                    }
                }
            }
            now=!now;
            pre=!pre;
            memset(dp[now],0,sizeof(dp[now]));
            for(int j=0;j<(1<<m);j++)
                for(int l=0;l<=d;l++){
                    dp[now][j<<1][l]=dp[pre][j][l];
                }
        }
        long long ans=0;
        for(int i=c;i<=d;i++)
            ans=(ans+dp[now][0][i])%mod;
        printf("%I64d\n",ans);
    }
    return 0;
}