Codeforces 225C Barcode

Test instructions: give you a matrix that contains only ' # ' and '. ', and now each column must be the same, up to a continuous y column with at least the same number of consecutive X columns.

Solution thinking: 3-D dp,dp[i][j][k] denotes column I, State J, is the K-line of State J.

Problem Solving Code:

1 //File name:225c.cpp2 //Author:darkdream3 //Created time:2015 March 09 Monday 09:49 05 Seconds4#include <climits>5#include <vector>6#include <list>7#include <map>8#include <Set>9#include <deque>Ten#include <stack> One#include <bitset> A#include <algorithm> -#include <functional> -#include <numeric> the#include <utility> -#include <sstream> -#include <iostream> -#include <iomanip> +#include <cstdio> -#include <cmath> +#include <cstdlib> A#include <cstring> at#include <ctime> - #defineLL Long Long - using namespacestd; - Charstr[1005][1005]; - intdp[1005][2][1005]; - intcw[1005]; in intBb:1005]; - intMain () { to   intN, m, x, y; +memset (dp,-1,sizeof(DP)); -scanf"%d %d%d%d",&n,&m,&x,&y); the    for(inti =1; I <= N;i + +) *   { $scanf"%s", &str[i][1]);Panax Notoginseng   } -    for(inti =1; I <= M;i + +) the   { +      for(intj =1; J <= N;j + +) A     { the       if(Str[j][i] = ='#') +Cw[i] + + ; -     } $Cb[i] = n-Cw[i]; $   } -dp[1][0][1] = cb[1]; -dp[1][1][1] = cw[1]; the    for(intI=2; I <= M;i + +) -   {Wuyi      intMXCB =-1; the       for(intj =1; J <= y; j + +) -      { Wu           if(dp[i-1][0][J]! =-1) -           { Aboutdp[i][0][j+1] = dp[i-1][0][J] +Cb[i]; $             if(J >=x) -             { -              if(MXCB = =-1) -MXCB = dp[i-1][0][j]; A              Else  +MXCB = min (mxcb,dp[i-1][0][j]); the             } -           } $      } the      intMXCW =-1;  the       for(intj =1; J <= y; j + +) the      { the           if(dp[i-1][1][J]! =-1) -           { indp[i][1][j+1] = dp[i-1][1][J] +Cw[i]; the             if(J >=x) the             { About              if(MXCW = =-1) theMXCW = dp[i-1][1][j]; the              Else  theMXCW = min (mxcw,dp[i-1][1][j]); +             } -           } the      }Bayi      //printf ("%d%d\n", MXCW,MXCB); the      if(MXCW! =-1) thedp[i][0][1] = MXCW +Cb[i]; -      if(MXCB! =-1) -dp[i][1][1] = MXCB +Cw[i]; the   } the   intAns =Int_max; the    for(inti =0; I <2; i + +) the        for(intj = x; J <= y; J + +) -       { the         if(Dp[m][i][j]! =-1) theAns =min (dp[m][i][j],ans); the       }94   /*for (int i = 1;i <= m;i + +) the   { the For (int s = 0; s <= 1; s + +) the    {98 for (int j= 1;j <= y; j + +) About printf ("%d", dp[i][s][j]); - printf ("* * *");101    }102 printf ("\ n");103   }*/104printf"%d\n", ans); the return 0;106}

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Codeforces 225C Barcode