HDU 4721 Food and Productivity (二分+树状数组)

转载请注明出处，谢谢http://blog.csdn.net/ACM_cxlove?viewmode=contents    by---cxlove

题意 ：给出n * m的格子，每个格子有两个属性food , prod。对于每一个查询A，B，可以选择某个格子将food属性+A，prod+B，然后以这个格子为中心的正方形两个属性和的最小值最大。

http://acm.hdu.edu.cn/showproblem.php?pid=4721

其实很简单的题， 很早就会了，WA了好多天，结果是一个乘号打成了加号。。。不能更逗

由于所有属性值为正，所以没必要取边界上不完整的正方形，不过处理下也没事。。。

正方形的连长为2 * r + 1，预处理一下子矩阵和。

将所有的正方形两个属性和提取出来，离散化一下food，用BIT维护prod属性的前缀最大值。

对于每一次查询，二分答案ans , 两个属性和都要大于等于ans。

二分下food，查询一下prod的最值。

逆序后维护前缀最大值。。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <set>#include <map>#include <vector>#include <queue>#include <stack>#define lson step << 1#define rson step << 1 | 1#define lowbit(x) (x & (-x))#define Key_value ch[ch[root][1]][0] using namespace std;typedef long long LL;const int N = 505;int n , m , r , q;int food[N][N] , prod[N][N];int a[N * N] , b[N * N] , tot , x[N * N];int s[N * N] , cnt;void add (int x , int v) {    for (int i = x ; i <= cnt ; i += lowbit (i)) {        s[i] = max (s[i] , v);    }}int ask (int x) {    int ret = 0;    for (int i = x ; i > 0 ; i -= lowbit (i)) {        ret = max (ret , s[i]);    }    return ret;}int main () {    #ifndef ONLINE_JUDGE        freopen ("input.txt" , "r" , stdin);        // freopen ("output.txt" , "w" , stdout);    #endif    int t , cas = 0;    scanf ("%d" , &t);    while (t --) {        memset (food , 0 , sizeof(food));        memset (prod , 0 , sizeof(prod));        memset (s , 0 , sizeof(s));        scanf ("%d %d %d %d" , &n , &m , &r , &q);        for (int i = 1 ; i <= n ; i ++) {            for (int j = 1 ; j <= m ; j ++) {                scanf ("%d" , &food[i][j]);                food[i][j] += food[i - 1][j] + food[i][j - 1] - food[i - 1][j - 1];            }        }        for (int i = 1 ; i <= n ; i ++) {            for (int j = 1 ; j <= m ; j ++) {                scanf ("%d" , &prod[i][j]);                prod[i][j] += prod[i - 1][j] + prod[i][j - 1] - prod[i - 1][j - 1];            }        }        tot = 0;        r = 2 * r + 1;        for (int i = r ; i <= n ; i ++) {            for (int j = r ; j <= m ; j ++) {                a[tot] = food[i][j] - food[i][j - r] - food[i - r][j] + food[i - r][j - r];                b[tot] = prod[i][j] - prod[i][j - r] - prod[i - r][j] + prod[i - r][j - r];                x[tot] = a[tot];                tot ++;            }        }        sort (x , x + tot);        cnt = unique (x , x + tot) - x;        for (int i = 0 ; i < tot ; i ++) {            int y = cnt - (lower_bound (x , x + cnt , a[i]) - x);            add (y , b[i]);        }        printf ("Case #%d:\n" , ++ cas);        while (q --) {            int A , B;            scanf ("%d %d" , &A , &B);            int low = min (A , B) + r * r * 1 , high = min (A , B) + r * r * 3 , mid , ans;            while (low <= high) {                mid = (low + high) >> 1;                int y = cnt - (lower_bound (x , x + cnt , mid - A) - x);                if (y < 1) {                    high = mid - 1;                    continue;                }                int tmp = ask (y);                if (tmp + B >= mid) {                    ans = mid;                    low = mid + 1;                }                else high = mid - 1;            }            printf ("%d\n" , ans);        }        if (t) puts ("");    }    return 0;}