Mosaic
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=95149#problem/G
Description
Can you help the God of sheep?
Input
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 action, print the new color value of the updated cell.
Sample Input
1 3 1 2 3 4 5 6 7 8 9 5 2 2 1 3 2 3 1 1 3 1 2 3 2 2 3
Sample Output
HINT
题意
给你一个n*n的矩阵，每次操作询问一个区域的(Max+Min)/2是多少
并且把这个区域的值全部改为(Max+Min)/2这个
题解：
二维线段树就好啦
代码：
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MinN = ;
struct Nodey
{
int l,r;
int Min,Max;
};
int locy[MinN],locx[MinN] , n , m, q; struct Nodex
{
int l,r;
Nodey sty[MinN*];
void build(int i,int _l,int _r)
{
sty[i].l = _l;
sty[i].r = _r;
sty[i].Min = sty[i].Max = ;
if(_l == _r)
{
locy[_l] = i;
return;
}
int mid = (_l + _r)/;
build(i<<,_l,mid);
build((i<<)|,mid+,_r);
}
int queryMin(int i,int _l,int _r)
{
if(sty[i].l == _l && sty[i].r == _r)
return sty[i].Min;
int mid = (sty[i].l + sty[i].r)/;
if(_r <= mid)return queryMin(i<<,_l,_r);
else if(_l > mid)return queryMin((i<<)|,_l,_r);
else return min(queryMin(i<<,_l,mid) , queryMin((i<<)|,mid+,_r));
}
int queryMax(int i,int _l,int _r)
{
if(sty[i].l == _l && sty[i].r == _r)
return sty[i].Max;
int mid = (sty[i].l + sty[i].r)/;
if(_r <= mid)return queryMax(i<<,_l,_r);
else if(_l > mid)return queryMax((i<<)|,_l,_r);
else return max(queryMax(i<<,_l,mid) , queryMax((i<<)|,mid+,_r));
}
}stx[MinN*]; void build(int i,int l,int r)
{
stx[i].l = l;
stx[i].r = r;
stx[i].build(,,);
if(l == r)
{
locx[l] = i;
return;
}
int mid = (l+r)/;
build(i<<,l,mid);
build((i<<)|,mid+,r);
}
//修改值
void Modify(int x,int y,int val)
{
int tx = locx[x];
int ty = locy[y];
stx[tx].sty[ty].Min = stx[tx].sty[ty].Max = val;
for(int i = tx;i;i >>= )
for(int j = ty;j;j >>= )
{
if(i == tx && j == ty)continue;
if(j == ty)
{
stx[i].sty[j].Min = min(stx[i<<].sty[j].Min , stx[(i<<)|].sty[j].Min);
stx[i].sty[j].Max = max(stx[i<<].sty[j].Max , stx[(i<<)|].sty[j].Max);
}
else
{
stx[i].sty[j].Min = min(stx[i].sty[j<<].Min , stx[i].sty[(j<<)|].Min);
stx[i].sty[j].Max = max(stx[i].sty[j<<].Max , stx[i].sty[(j<<)|].Max);
}
}
}
int queryMax(int i,int x1,int x2,int y1,int y2)
{
if(stx[i].l == x1 && stx[i].r == x2)
return stx[i].queryMax(,y1,y2);
int mid = (stx[i].l + stx[i].r)/;
// cout << stx[i].l << " " << stx[i].r << " " << mid << endl;
if(x2 <= mid)return queryMax(i<<,x1,x2,y1,y2);
else if(x1 > mid)return queryMax((i<<)|,x1,x2,y1,y2);
else return max(queryMax(i<<,x1,mid,y1,y2) , queryMax((i<<)|,mid+,x2,y1,y2));
}
int queryMin(int i,int x1,int x2,int y1,int y2)
{
if(stx[i].l == x1 && stx[i].r == x2)
return stx[i].queryMin(,y1,y2);
int mid = (stx[i].l + stx[i].r)/;
// cout << stx[i].l << " " << stx[i].r << " " << mid << endl;
if(x2 <= mid)return queryMin(i<<,x1,x2,y1,y2);
else if(x1 > mid)return queryMin((i<<)|,x1,x2,y1,y2);
else return min(queryMin(i<<,x1,mid,y1,y2) , queryMin((i<<)|,mid+,x2,y1,y2));
} int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
scanf("%d",&T);int m;
for(int cas = ;cas <= T;cas++)
{
printf("Case #%d:\n",cas);
int q;
scanf("%d",&n); build(,,);
for(int i = ;i <= n;i++)
for(int j = ;j <= n;j++)
{
int a;
scanf("%d",&a);
Modify(i,j,a);
}
int x,y,L;
scanf("%d",&q);
while(q--)
{
scanf("%d%d%d",&x,&y,&L);
int x1 = max(x-L/,);
int y1 = max(y-L/,);
int x2 = min(x+L/,n);
int y2 = min(y+L/,n);
int t = queryMax(,x1,x2,y1,y2) + queryMin(,x1,x2,y1,y2);
//cout<<queryMax(1,x,x2,y,y2) <<" "<< queryMin(1,x,x2,y,y2) << endl;
t = t/;
printf("%d\n",t);
Modify(x,y,t);
}
}
return ;
}