Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

It's graduated season, every students should leave something on the wall, so....they draw a lot of geometry shape with different color.

When teacher come to see what happened, without getting angry, he was surprised by the talented achievement made by students. He found the wall full of color have a post-modern style so he want to have an in-depth research on it.

To simplify the problem, we divide the wall into n*m ( 1n200, 1m50000) pixels, and we have got the order of coming students who drawing on the wall. We found that all students draw four kinds of geometry shapes in total that is Diamond, Circle, Rectangle and Triangle. When a student draw a shape in pixel (i, j) with color c ( 1c9), no matter it is covered before, it will be covered by color c.

There are q ( 1q50000) students who have make a drawing one by one. And after q operation we want to know the amount of pixels covered by each color.

Input 

There are multiple test cases.

In the first line of each test case contains three integers n, m, q. The next q lines each line contains a string at first indicating the geometry shape:

• Circle: given xc, yc, r, c, and you should cover the pixels (x, y) which satisfied inequality (x - xc)² + (y - yc)²r² with color c;
• Diamond: given xc, yc, r, c, and you should cover the pixels (x, y) which satisfied inequality | x - xc| + | y - yc|r with color c;
• Rectangle: given xc, yc, l, w, c, and you should cover the pixels (x, y) which satisfied xcxxc + l - 1, ycyyc + w - 1 with color c;
• Triangle: given xc, yc, w, c, W is the bottom length and is odd, the pixel (xc, yc) is the middle of the bottom. We define this triangle is isosceles and the height of this triangle is (w + 1)/2, you should cover the correspond pixels with color c;

Note: all shape should not draw out of the n*m wall! You can get more details from the sample and hint. (0xc, xn - 1, 0yc, ym - 1)

Output 

For each test case you should output nine integers indicating the amount of pixels covered by each color.

Hint: The final distribution of different colors:

00000000
03300000
03310000
00111000
00022240
00002444
00004444
00000444

Sample Input 

 
8 8 4
Diamond 3 3 1 1
Triangle 4 4 3 2
Rectangle 1 1 2 2 3
Circle 6 6 2 4

Sample Output 

 
4 4 4 11 0 0 0 0 0

---

調整心理的模擬題。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<map>
#define mem(x) memset(x,0,sizeof(x))
using namespace std;

typedef long long ll;
char g[200][50000];
int main(){
    long long n, m, q, i, j, k;
    while(scanf("%lld %lld %lld", &n, &m, &q) == 3) {
        for(i = 0; i < n; i++)
            for(j = 0; j < m; j++)
                g[i][j] = 0;
        char cmd[100];
        long long x, y, r, c, l, w, W;
        while(q--) {
            scanf("%s", cmd);
            if(!strcmp(cmd, "Circle")) {
                scanf("%lld %lld %lld %lld", &x, &y, &r, &c);
                for(i = x-r; i <= x+r; i++) {
                    for(j = y-r; j <= y+r; j++) {
                        if(i < 0 || j < 0 || i >= n || j >= m)
                            continue;
                        if((i-x)*(i-x)+(j-y)*(j-y) > r*r)
                            continue;
                        g[i][j] = c;
                    }
                }
            } else if(!strcmp(cmd, "Diamond")) {
                scanf("%lld %lld %lld %lld", &x, &y, &r, &c);
                for(i = x-r; i <= x+r; i++) {
                    for(j = y-r; j <= y+r; j++) {
                        if(i < 0 || j < 0 || i >= n || j >= m)
                            continue;
                        if(abs(i-x)+abs(j-y) > r)
                            continue;
                        g[i][j] = c;
                    }
                }
            } else if(!strcmp(cmd, "Rectangle")) {
                scanf("%lld %lld %lld %lld %lld", &x, &y, &l, &w, &c);
                for(i = x; i < x+l; i++) {
                    for(j = y; j < y+w; j++) {
                        if(i < 0 || j < 0 || i >= n || j >= m)
                            continue;
                        g[i][j] = c;
                    }
                }
            } else if(!strcmp(cmd, "Triangle")) {
                scanf("%lld %lld %lld %lld", &x, &y, &w, &c);
                int tw = w;
                for(i = x; i <= x+(tw+1)/2+52; i++, w-=2) {
                    for(j = y-(w+1)/2+1; j <= y+(w+1)/2-1; j++) {
                        if(i < 0 || j < 0 || i >= n || j >= m)
                            continue;
                        g[i][j] = c;
                    }
                }
            }
        }
        int cnt[10] = {};
        for(i = 0; i < n; i++)
            for(j = 0; j < m; j++)
                cnt[g[i][j]]++;
        for(i = 1; i < 10; i++) {
            if(i != 1)
                printf(" ");
            printf("%d", cnt[i]);
        }
        puts("");
    }
    return 0;
}