Ultraviolet a 1493-Draw a Mess (and query set)

Ultraviolet a 1493-Draw a Mess (and query set)
Ultraviolet a 1493-Draw a Mess

Question Link

There are four operations on an n * m plane. The corresponding area is colored as v (1 <= v <= 9). After q operations, how many colors are required?

Train of Thought: Check the set, because the color coated will cover, so that we can perform the reverse operation, so that each operation can not be painted if there is a color before, if not, you can paint, then there are a total of 200 rows. Each row uses and queries the set compression path to find the next location that can be colored.

In the question, the triangle edge must be an odd number, but an even number exists... This pitfall.

Code:

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include using namespace std;const int N = 205;const int M = 50005;struct OP {    int c, xc, yc, a, b, v;    OP(){}    OP(int c, int xc, int yc, int a, int b, int v) {this->c = c; this->xc = xc; this->yc = yc;this->a = a; this->b = b; this->v = v;    }} op[M];int parent[N][M];int find(int *parent, int x) {    return parent[x] == x ? x : parent[x] = find(parent, parent[x]);}int n, m, q, ans[10];void init() {    for (int i = 0; i < n; i++)for (int j = 0; j <= m; j++)    parent[i][j] = j;    memset(ans, 0, sizeof(ans));    char str[15];    int a, b, c, d, e;    for (int i = 0; i < q; i++) {scanf("%s", str);scanf("%d%d%d%d", &a, &b, &c, &d);if (str[0] == 'C')    op[i] = OP(0, a, b, c, 0, d);if (str[0] == 'D')    op[i] = OP(1, a, b, c, 0, d);if (str[0] == 'R') {    scanf("%d", &e);    op[i] = OP(2, a, b, c, d, e);}if (str[0] == 'T')    op[i] = OP(3, a, b, c, 0, d);    }}void gao(int l, int r, int *parent, int v) {    l = find(parent, l);    while (l <= r) {ans[v]++;int tmp = find(parent, l + 1);parent[l] = tmp;l = tmp;    }}void cir(int xc, int yc, int r, int v) {    for (int i = max(0, xc - r); i <= min(n - 1, xc + r); i++) {int len = (int)(sqrt((r * r - (i - xc) * (i - xc))));int l = max(0, yc - len), r = min(m - 1, yc + len);gao(l, r, parent[i], v);    }}void dia(int xc, int yc , int r, int v) {    for (int i = max(0, xc - r); i <= min(n - 1, xc + r); i++) {int len = r - abs(i - xc);int l = max(0, yc - len), r = min(m - 1, yc + len);gao(l, r, parent[i], v);    }}void rec(int xc, int yc, int h, int w, int v) {    for (int i = max(0, xc); i <= min(n - 1, xc + h - 1); i++) {int l = max(0, yc), r = min(m - 1, yc + w - 1);gao(l, r, parent[i], v);    }}void tri(int xc, int yc, int w, int v) {    for (int i = max(0, xc); i <= min(n - 1, xc + (w + 1) / 2 - 1); i++) {int len = (w - 1) / 2 - i + xc;int l = max(0, yc - len), r = min(m - 1, yc + len);gao(l, r, parent[i], v);    }}void solve() {    for (int i = q - 1; i >= 0; i--) {if (op[i].c == 0)    cir(op[i].xc, op[i].yc, op[i].a, op[i].v);if (op[i].c == 1)    dia(op[i].xc, op[i].yc, op[i].a, op[i].v);if (op[i].c == 2)    rec(op[i].xc, op[i].yc, op[i].a, op[i].b, op[i].v);if (op[i].c == 3)    tri(op[i].xc, op[i].yc, op[i].a, op[i].v);    }    for (int i = 1; i <= 9; i++)printf("%d%c", ans[i], i == 9 ? '\n' : ' ');}int main() {    while (~scanf("%d%d%d", &n, &m, &q)) {init();solve();    }    return 0;}