ZOJ 3347 Picture Handling

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http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3347

Xiaoyao 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:

1. Invert: For any pixel with value v in selected area, change v to -v.
2. Lighten: For any pixel with value v in selected area, increase v by 1.
3. Darken: For any pixel with value v in selected area, decrease v by 1.
4. Flip Horizontal: For any pixel at (x, y) in selected area, replace its value with pixel at (x₁ + x₂ - x, y).
5. 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;
}

ZOJ 3347 Picture Handling

标签：

原文地址：http://blog.csdn.net/hellohelloc/article/details/47834617