[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 = 100010;
int n;
char s[maxn];
ll grid[maxn];

int main() {
    // freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    for(int _ = 1; _ <= T; _++) {
        scanf("%s", s);
        memset(grid, 0, sizeof(grid));
        ll cur = 0, ans = 0;
        n = 0; grid[n] = 1;
        for(int i = 1; s[i]; i++) s[i] == s[i-1] ? grid[n]++ : grid[++n] = 1;
        for(int i = 0; i <= n; i++)    cur += grid[i] * grid[i];
        if(n == 0) ans = grid[0] * grid[0];
        else {
            for(int i = 1; i <= n; i++) {
                if(grid[i] == 1) ans = max(cur+2*(grid[i-1]*grid[i+1]+grid[i-1]+grid[i+1]), ans);
                else {
                    if(grid[i-1] >= grid[i]) 
                        ans = max(cur+2*(grid[i-1]-grid[i]+1), ans);
                    else 
                        ans = max(cur+2*(grid[i]-grid[i-1]+1), ans);
                }
            }
        }
        printf("Case #%d: %I64d
", _, ans);
    }
    return 0;
}