HDU 4819 Mosaic 二维线段树

二维RMQ问题

用二维线段树可以实现

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <queue>#include <map>#include <string>#include <stack>#define maxn 5000#define _max(a,b) ((a)>(b)?(a):(b))#define _min(a,b) ((a)>(b)?(b):(a))const int maxi = 1 << 30;const int mini = 1 << 31;const int inf = 0x3f3f3f3f;using namespace std;struct SegTree {    int maxx[maxn][maxn], minn[maxn][maxn], n, m;    int xo, x1, x2, y1, y2, resmin, resmax, x, y, val, isleaf;    void query1D(int o, int l, int r) {        if(y1 <= l && r <= y2) {            resmin = min(resmin, minn[xo][o]);            resmax = max(resmax, maxx[xo][o]);            //printf("y1:%d y2:%d l:%d r:%d %d %d\n", y1, y2, l, r, minn[xo][o], maxx[xo][o]);        }        else {            int mid = l + (r - l) / 2;            if(y1 <= mid) query1D(2 * o, l, mid);            if(mid < y2) query1D(2 * o + 1, mid + 1, r);        }    }    void query2D(int o, int l, int r) {        if(x1 <= l && x2 >= r) {            xo = o;            query1D(1, 1, m);        }        else {            int mid = l + (r - l) / 2;            if(x1 <= mid) query2D(2 * o, l, mid);            if(x2 > mid) query2D(2 * o + 1, mid + 1, r);        }    }    void change1D(int o, int l, int r) {        if(l == r) {            if(isleaf) {                maxx[xo][o] = minn[xo][o] = val;                return ;            }            maxx[xo][o] = max(maxx[2 * xo][o], maxx[2 * xo + 1][o]);            minn[xo][o] = min(minn[2 * xo][o], minn[2 * xo + 1][o]);        }        else {            int mid = l + (r - l) / 2;            if(y <= mid) change1D(2 * o, l, mid);            else change1D(2 * o + 1, mid + 1, r);            maxx[xo][o] = max(maxx[xo][2 * o], maxx[xo][2 * o + 1]);            minn[xo][o] = min(minn[xo][2 * o], minn[xo][2 * o + 1]);        }    }    void change2D(int o, int l, int r) {        if(l == r) {            xo = o;            isleaf = 1;            change1D(1, 1, m);        }        else {            int mid = l + (r - l) / 2;            if(x <= mid) change2D(2 * o, l, mid);            else change2D(2 * o + 1, mid + 1, r);            xo = o;            isleaf = 0;            change1D(1, 1, m);        }    }    void query() {        resmin = inf;        resmax = -inf;        query2D(1, 1, n);    }    void change() {        change2D(1, 1, n);    }};SegTree Tree;int main() {    #ifdef LOCAL    freopen("in.txt", "r", stdin);    #endif    int T, cas = 1;    scanf("%d", &T);    while(T--) {        printf("Case #%d:\n", cas++);        int n;        scanf("%d", &n);        Tree.n = Tree.m = n;        for(int i = 1; i <= n; i++) {            for(int j = 1; j <= n; j++) {                int t;                scanf("%d", &t);                Tree.x = i, Tree.y = j, Tree.val = t;                Tree.change();            }        }        int m;        scanf("%d", &m);        for(int i = 0; i < m; i++) {            int a, b, c;            scanf("%d%d%d", &a, &b, &c);            Tree.x1 = max(1, a - c / 2);            Tree.x2 = min(n, a + c / 2);            Tree.y1 = max(1, b - c / 2);            Tree.y2 = min(n, b + c / 2);            Tree.query();            Tree.val = (Tree.resmin + Tree.resmax) / 2;            //printf("%d %d ", Tree.resmin, Tree.resmax);            printf("%d\n", Tree.val);            Tree.x = a, Tree.y = b;            Tree.change();        }    }    return 0;}

题目:

Mosaic

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 685    Accepted Submission(s): 266

Problem Description

The God of sheep decides to pixelate some pictures (i.e., change them into pictures with mosaic). Here's how he is gonna make it: for each picture, he divides the picture into n x n cells, where each cell is assigned a color value. Then he chooses a cell, and checks the color values in the L x L region whose center is at this specific cell. Assuming the maximum and minimum color values in the region is A and B respectively, he will replace the color value in the chosen cell with floor((A + B) / 2).

Can you help the God of sheep?

 

Input

The first line contains an integer T (T ≤ 5) indicating the number of test cases. Then T test cases follow.

Each test case begins with an integer n (5 < n < 800). Then the following n rows describe the picture to pixelate, where each row has n integers representing the original color values. The j-th integer in the i-th row is the color value of cell (i, j) of the picture. Color values are nonnegative integers and will not exceed 1,000,000,000 (10^9).

After the description of the picture, there is an integer Q (Q ≤ 100000 (10^5)), indicating the number of mosaics.

Then Q actions follow: the i-th row gives the i-th replacement made by the God of sheep: xi, yi, Li (1 ≤ xi, yi ≤ n, 1 ≤ Li < 10000, Li is odd). This means the God of sheep will change the color value in (xi, yi) (located at row xi and column yi) according to the Li x Li region as described above. For example, an query (2, 3, 3) means changing the color value of the cell at the second row and the third column according to region (1, 2) (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4). Notice that if the region is not entirely inside the picture, only cells that are both in the region and the picture are considered.

Note that the God of sheep will do the replacement one by one in the order given in the input.

 

Output

For each test case, print a line "Case #t:"(without quotes, t means the index of the test case) at the beginning.

For each action, print the new color value of the updated cell.

 

Sample Input

131 2 34 5 67 8 952 2 13 2 31 1 31 2 32 2 3

 

Sample Output

Case #1:56346

 

Source

2013 Asia Regional Changchun