Description
After years as a brick-layer, you've been called upon to analyze the instability of brick walls. The instability of a wall can be approximated by the maximum damage to a wall in case of taking one brick out. A brick will fall if all bricks that are directly underneath it are removed. Note that if the space underneath a brick is partially empty, it does not fall. You are given the description of all bricks in a wall, and must determine the instability of the wall as described in the following sections.
Input
There are multiple test cases in the input. Each test case consists of a single line, ``M N " (1
M, N100) where M and N indicate the height and width (in units), respectively, of the input wall.
Each of the next M lines is a string of N digits which specifies a row in the wall. Each brick in a row is represented by a substring of the row (like s
) such that every digit in s is the same, which is equal to the length of s
too. For example, 333 and 22 are two bricks of length 3 and 2
respectively, but 111 specifies three bricks of length one. A 0 in a row
means there is no brick in that place of wall. Note that the height of
each brick is one. The input terminates with a line containing `
0 0'. You may assume that the input is correct. This means:
- There is no brick such that the length of the brick does not conform to the digits in the brick (like 222 in the row 12221).
- No brick can fall initially.
Output
For each test case, write a single line containing maximum sum of the bricks' lengths that will fall if we take one brick out (including that brick).
Sample Input
4 5 33322 22333 33322 22333 4 6 122333 444422 111111 333333 3 3 022 220 111 0 0
Sample Output
5 8 4
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 100001 const int inf=0x7fffffff; //无限大 int dp[110][110]; int g[110][110]; int q[110][110]; int main() { int n,m; string s[110]; while(cin>>n>>m) { if(n==0&&m==0) break; memset(dp,0,sizeof(dp)); memset(q,0,sizeof(q)); memset(g,0,sizeof(g)); for(int i=0;i<n;i++) { cin>>s[i]; for(int j=0;j<m;j++) { if(s[i][j]=='0') { g[i][j]=-1; } else { g[i][j]=(int)(s[i][j]-'0'); j=j+(int)(s[i][j]-'0')-1; } } } for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { if(i==0) { dp[i][j]=(int)(s[i][j]-'0'); } else { if(g[i][j]==-1) dp[i][j]=0; else if(g[i][j]==0) dp[i][j]=dp[i][j-1]; else { int ii=i-1; int sum=0; int op=g[i][j]; dp[i][j]+=g[i][j]; for(int k=0;k<m;k++) { if(j<=k&&k<j+op) q[i][k]=-1; else q[i][k]=g[i][k]; } while(1) { sum=0; for(int k=0;k<m;k++) { if(q[ii][k]==-1) continue; q[ii][k]=g[ii][k]; for(int jj=0;jj<g[ii][k];jj++) { if(q[ii+1][k+jj]!=-1) break; if(jj==g[ii][k]-1) { for(int mm=0;mm<g[ii][k];mm++) { q[ii][k+mm]=-1; sum++; } } } } dp[i][j]+=sum; if(sum==0) break; ii--; if(ii<0) break; } memset(q,0,sizeof(q)); } } } } int ans=0; for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { ans=max(ans,dp[i][j]); } } cout<<ans<<endl; } return 0; }
