ZOJ 3614 Choir

东西学的多了不一定好，因为理解的不够深入，所以会乱用。比如这道题目。。。。

学了线段树后知道线段树也可以求第k大值，结果TLE。。。

学了树状数组后知道树状数组也可以求第k大值，结果又TLE。。。

实在优化不过去了，看看被人怎么写的？RMQ啊！

求区间内的最值用RMQ，O(1)的查询，怎么把这个忘了。。。

越学就越不知道用哪个好了。。。

还要加深理解。。。

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>using namespace std;#define LL long longconst 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;}