HDOJ City hall 1453(模拟)

City hall

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 256    Accepted Submission(s): 160

Problem Description

Because of its age, the City Hall has suffered damage to one of its walls. A matrix with M rows and N columns represents the encoded image of that wall, where 1 represents an intact wall and 0 represents a damaged wall (like in Figure-1).

               1110000111
               1100001111
               1000000011
               1111101111
               1110000111

                   Figure-1

To repair the wall, the workers will place some blocks vertically into the damaged area. They can use blocks with a fixed width of 1 and different heights of {1,2, ..., M}. 

For a given image of the City Hallˇs wall, your task is to determine how many blocks of different heights are needed to fill in the damaged area of the wall, and to use the least amount of blocks.

 

Input

There is only one test case. The case starts with a line containing two integers M and N (1 <= M, N <= 200). Each of the following M lines contains a string with length of N, which consists of ¨1〃s and/or ¨0〃s. These M lines represent the wall.

 

Output

You should output how many blocks of different heights are needed. Use separate lines of the following format:

k Ck

where k劇{1,2, ..., M} means the height of the block, and Ck means the amount of blocks of height k that are needed. You should not output the lines where Ck = 0. The order of lines is in the ascending order of k.

 

Sample Input

5 1011100001111100001111100000001111111011111110000111

 

Sample Output

1 72 13 25 1

 

Source

ACM暑期集训队练习赛（七）

 

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#include<stdio.h>#include<string.h>#include<algorithm>#define INF 0x3f3f3f3fusing namespace std;int main(){double l;while(scanf("%lf",&l)!=EOF){int n;double c,t;double dp[110];double p[110];memset(p,0,sizeof(p));memset(dp,0,sizeof(dp));scanf("%d%lf%lf",&n,&c,&t);double vr,v1,v2;scanf("%lf%lf%lf",&vr,&v1,&v2);for(int i=1;i<=n;i++)scanf("%lf",&p[i]);p[0]=0;p[n+1]=l;double T1=l/vr,time;dp[0]=0;for(int i=1;i<=n+1;i++){dp[i]=INF;for(int j=0;j<i;j++){double S=p[i]-p[j];if(S>c)time=(c/v1)+(S-c)/v2;else time=S/v1;if(j!=0)time+=t;dp[i]=min(dp[i],dp[j]+time);}}if(T1>dp[n+1])printf("What a pity rabbit!\n");elseprintf("Good job,rabbit!\n");}return 0;}