ZOJ 3614 Choir

東西學的多了不一定好，因爲理解的不夠深入，所以會亂用。比如這道題目。。。。

學了線段樹後知道線段樹也可以求第k大值，結果TLE。。。

學了樹狀數組後知道樹狀數組也可以求第k大值，結果又TLE。。。

實在優化不過去了，看看被人怎麼寫的？RMQ啊！

求區間內的最值用RMQ，O(1)的查詢，怎麼把這個忘了。。。

越學就越不知道用哪個好了。。。

還要加深理解。。。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
#define LL long long
const double eps = 1e-9;
const double INF = 1e10;
const int maxn = 310;
int sum[maxn][maxn];
LL squ[maxn][maxn];
int mx[10][10][maxn][maxn];
int m, n, a, b, x, y;
double ans;
void rmqinit()
{
    for(int r = 0; (1 << r) < m; r++){
        for(int c = 0; (1 << c) < n; c++){
            if(r == 0 && c == 0) continue;
            for(int i = m; i >= 1; i--){
                for(int j = n; j >= 1; j--){
                    if(r == 0){
                        mx[r][c][i][j] = mx[r][c - 1][i][j];
                        if(j + (1 << (c - 1)) <= n)
                            mx[r][c][i][j] = max(mx[r][c][i][j], mx[r][c - 1][i][j + (1 << (c - 1))]);
                    }else {
                        mx[r][c][i][j] = mx[r - 1][c][i][j];
                        if(i + (1 << (r - 1)) <= m)
                            mx[r][c][i][j] = max(mx[r][c][i][j], mx[r - 1][c][i + (1 << (r - 1))][j]);
                    }
                }
            }
        }
    }
}
int rmq(int r1, int c1, int r2, int c2)
{
    int m1 = max((int)floor(log((double)(r2 - r1 + 1)) / log(2.0) - eps), 0);
    int m2 = max((int)floor(log((double)(c2 - c1 + 1)) / log(2.0) - eps), 0);
    return max(max(mx[m1][m2][r1][c1], mx[m1][m2][r1][c2 - (1 << m2) + 1]), max(mx[m1][m2][r2 - (1 << m1) + 1][c1], mx[m1][m2][r2 - (1 << m1) + 1][c2 - (1 << m2) + 1]));
}

inline void slove(int u, int i, int j)
{
    int _sum = sum[i][j] - u;
    LL _squ = squ[i][j] - u * u;
    if(i + a <= m){
        _sum -= sum[i + a][j];
        _squ -= squ[i + a][j];
    }
    if(j + b <= n){
        _sum -= sum[i][j + b];
        _squ -= squ[i][j + b];
    }
    if(i + a <= m && j + b <= n){
        _sum += sum[i + a][j + b];
        _squ += squ[i + a][j + b];
    }
    int c = a * b - 1;
    double avg = (double)_sum / c;
    double v = (double)_squ / c + avg * avg - 2.0 * avg * _sum  / c;
    if(ans > v){
        ans = v;
        x = i;
        y = j;
    }
}

int main()
{
    //freopen("input.txt", "r", stdin);
    int q, con = 1;
    while(scanf("%d %d", &m, &n) == 2){
        printf("Case %d:\n", con++);
        for(int i = 1; i <= m; i++)
            for(int j = 1; j <= n; j++)
                scanf("%d", &mx[0][0][i][j]);
        for(int j = m; j >= 1; j--){
            int _sum = 0;
            LL _squ = 0;
            for(int i = n; i >= 1; i--){
                if(j < m) {
                    sum[j][i] = sum[j + 1][i];
                    squ[j][i] = squ[j + 1][i];
                }else {
                    sum[j][i] = 0; squ[j][i] = 0;
                }
                _sum += mx[0][0][j][i];
                _squ += mx[0][0][j][i] * mx[0][0][j][i];
                sum[j][i] += _sum;
                squ[j][i] += _squ;
            }
        }
        rmqinit();
        scanf("%d", &q);
        while(q--){
            scanf("%d %d", &a, &b);
            ans = INF;
            for(int i = 1; i + a - 1 <= m; i++){
                for(int j = 1; j + b - 1 <= n; j++){
                    int u = rmq(i, j, i + a - 1, j + b - 1);
                    slove(u, i, j);
                }
            }
            printf("(%d, %d), %.2lf\n", x, y, ans);
        }
    }
    return 0;
}