[置顶] ZOJ 3347 Picture Handling
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3347
Picture Handling Time Limit: 1 Second Memory Limit: 32768 KBXiaoyao likes to play with pictures very much. When he got a picture, he will use rectangle selection tool to select an area ((x1, y1) to (x2, y2), inclusively) and perform these operations:
- Invert: For any pixel with value v in selected area, change v to -v.
- Lighten: For any pixel with value v in selected area, increase v by 1.
- Darken: For any pixel with value v in selected area, decrease v by 1.
- Flip Horizontal: For any pixel at (x, y) in selected area, replace its value with pixel at (x1 + x2 - x, y).
- Flip Vertical: For any pixel at (x, y) in selected area, replace its value with pixel at (x, y1 + y2 - y).
After several operations, Xiaoyao wonders what value a pixel at specified position is. Could you tell him?
Input
There are multi cases (no more than 5). Please proceed to the end of input. Each case is like below:
The first line contains two integers W and H, indicating the width and height of the picture. W and H are both between 1 and 255, inclusively.
Following H lines, each line contains W integers, indicating the value of pixels. The first integer of the first line in these H lines is the value of pixel at (0, 0) and the last integer is the value of pixel at (W-1, H-1). All these values are between -105 and 105, inclusively.
Then, a line with one integer M(0 <= M <= 105).
Following M lines, each line contains 5 integers: x1 y1 x2 y2 op, indicating the selected area and the operation. (0 <= x1 <= x2 < W, 0 <= y1 <= y2 < H, 1 <= op <= 5). If op is 1, it means Invert operation. If op is 2, it means Lighten operation. The others follow by analogy. You should follow the order of the input to perform operations.
Finally, there is a line with two integers x and y. (0 <= x < W, 0 <= y < H) Your task is to output the value of pixel at (x, y) after performing above operations.
Output
For each case, output a single line containing one integer which is the value of pixel at (x, y) at last.
Sample Input
3 2
1 2 3
4 5 6
2
0 0 1 1 1
1 0 2 1 4
2 1
Sample Output
-5
题意：给你一系列对矩阵子阵的操作，包括negative, –, ++, 水平翻转和垂直翻转。最后问某个位置的元素的值。
分析：由于只问一个位置的元素的值，所以只要简单的对其进行模拟操作就可以了，注意这个位置是最终位置，不是初始位置，所以操作的时候要逆序。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=100010;
struct node{
int op;
int x1,y1;
int x2,y2;
}t[maxn];
int a[300][300];
int m,h,k;
int main()
{
int x,y;
while(scanf("%d %d",&m,&h) != EOF)
{
memset(t,0,sizeof(t));
for(int i=0;i<h;i++)
for(int j=0;j<m;j++)
scanf("%d",&a[i][j]);
scanf("%d",&k);
for(int i=1;i<=k;i++)
scanf("%d %d %d %d %d",&t[i].x1,&t[i].y1,&t[i].x2,&t[i].y2,&t[i].op);
scanf("%d %d",&x,&y);
int fu=1;
int num=0;
for(int i=k;i>=1;i--)
{
if(x>=t[i].x1 && x<=t[i].x2 && y>=t[i].y1 && y<=t[i].y2){
if(t[i].op==1)
{
fu = -fu;
num = -num;
}
else if(t[i].op==2)
{
num++;
}
else if(t[i].op==3){
num--;
}
else if(t[i].op==4){
x=t[i].x1+t[i].x2-x;
}
else{
y=t[i].y1+t[i].y2-y;
}
}
}
printf("%d\n",fu*(a[y][x]+num));
}
return 0;
}