[HDOJ5583]Kingdom of Black and White(暴力)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5583

一个01串，求修改一个位置，使得所有数均为0或1的子串长度的平方和最大。先分块，然后统计好原来的结果，每次更新块，更新最大值。

 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>

 using namespace std;

 typedef long long ll;
 const int maxn = ;
 int n;
 char s[maxn];
 ll grid[maxn];

 int main() {
     // freopen("in", "r", stdin);
     int T;
     scanf("%d", &T);
     for(int _ = ; _ <= T; _++) {
         scanf("%s", s);
         memset(grid, , sizeof(grid));
         ll cur = , ans = ;
         n = ; grid[n] = ;
         for(int i = ; s[i]; i++) s[i] == s[i-] ? grid[n]++ : grid[++n] = ;
         for(int i = ; i <= n; i++)    cur += grid[i] * grid[i];
         if(n == ) ans = grid[] * grid[];
         else {
             for(int i = ; i <= n; i++) {
                 if(grid[i] == ) ans = max(cur+*(grid[i-]*grid[i+]+grid[i-]+grid[i+]), ans);
                 else {
                     if(grid[i-] >= grid[i])
                         ans = max(cur+*(grid[i-]-grid[i]+), ans);
                     else
                         ans = max(cur+*(grid[i]-grid[i-]+), ans);
                 }
             }
         }
         printf("Case #%d: %I64d\n", _, ans);
     }
     return ;
 }