Codeforces Round #601 (Div. 2) D. Feeding Chicken

Topic Address: http://codeforces.com/contest/1255/problem/D

Meaning of the questions: k chickens to decimeter, there are some places where there is no rice rice, chicken all points of difference to the smallest, to even up the score can be regarded as a collection.

Idea: to traverse the map again, the key is to go bow-shaped, so we can guarantee to each chicken together. Provided m total sum, the bird can be divided sum / k to block has m. sum% k of sharing it wants (see here to understand the code).

AC Code:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 typedef long long ll;
 7 char s[63];
 8 void sol(){
 9     int r,c,k;
10     scanf("%d%d%d",&r,&c,&k);
11     char maps[101][101];
12     for(int i=0;i<r;i++)
13         scanf("%s",maps[i]);
14     int sum=0;
15     for(int i=0;i<r;i++)
16         for(int j=0;j<c;j++)
17             if(maps[i][j]=='R') sum++;
18     int a=0,b=0;
19     for(int i=0;i<r;i++){
20 is          IF (I & . 1 ) {
 21 is              for ( int J = 0 ; J <C; J ++ ) {
 22 is                  IF (B + (Maps [I] [J] == ' R & lt ' )> ((A <SUM% K)? (SUM / K + . 1 ) :( SUM / K))) {       // first partial sum% k th sum / k + 1, the remaining sub-SUM / K 
23 is                      a ++ ;
 24                      Maps [I] [J] = S [ A];
 25                      B = . 1 ;
 26 is                  }
 27                  the else {
 28                      B = + (Maps [I] [J] == ' R & lt' );    // for each assigned to a recorded bit with a meter 
29                      Maps [I] [J] = S [A];
 30                  }
 31 is              }
 32          }
 33 is          the else {
 34 is              for ( int J = the C- . 1 ; J> = 0 ; J, ) {
 35                  IF (B + (Maps [I] [J] == ' R & lt ' )> (? (A <SUM% K) (SUM / K + . 1 ) :( SUM / K))) {
 36                      A ++ ;
 37 [                      Maps [I] [J] = S [A];
 38 is                     b=1;
39                 }
40                 else{
41                     b+=(maps[i][j]=='R');
42                     maps[i][j]=s[a];
43                 }
44             }
45         }
46     }
47     for(int i=0;i<r;i++){
48         for(int j=0;j<c;j++)
49             printf("%c",maps[i][j]);
50         printf("\n");
51     }
52 }
53 int main(){
54     int T;
55     cin>>T;
56     for(int i=0;i<62;i++){
57         if(i<26) s[i]=i+'a';
58         else if(i<52) s[i]=i-26+'A';
59         else s[i]=i-52+'0';
60     }
61     while(T--) sol();
62     return 0;
63 }