题目链接：uva 1493 - Draw a Mess

题目大意：给定一个矩形范围，有四种上色方式，后面上色回将前面的颜色覆盖，最后问9种颜色各占多少的区域。

解题思路：用并查集维护每个位置对应下一个可以上色的位置。然后将上色倒转过来处理，就解决了颜色覆盖的问题。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>

using namespace std;
const int maxr = 205;
const int maxc = 50005;

int N, M, Q, cnt[15];
int f[maxr][maxc];

struct Order {
    int sign, star, end;
    int x, y, r, w, c;
    void set (int sign, int x, int y, int c, int r, int w = 0) {
        this->sign = sign;
        this->x = x;
        this->y = y;
        this->c = c;
        this->r = r;
        this->w = w;
        del_star();
    }

    void del_star () {
        if (sign < 2) {
            star = max(x - r, 0);
            end = min(x + r, N - 1);
        } else if (sign == 2) {
            r = (r + 1) / 2 - 1;
            star = x;
            end = min(x + r, N-1);
        }
    }

}op[maxc];


int getfar (int* far, int x) {
    return x == far[x] ? x : far[x] = getfar(far, far[x]);
}

int style (char ch) {
    if (ch == ‘C‘)
        return 0;
    else if (ch == ‘D‘)
        return 1;
    else if (ch == ‘T‘)
        return 2;
    else
        return 3;
}

void init () {
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i <= M; i++) {
        for (int j = 0; j < N; j++)
            f[j][i] = i;
    }

    char s[20];
    int x, y, r, c, w;
    for (int i = 1; i <= Q; i++) {
        scanf("%s", s);
        if (s[0] != ‘R‘) {
            scanf("%d%d%d%d", &x, &y, &r, &c);
            op[i].set(style(s[0]), x, y, c, r);
        } else {
            scanf("%d%d%d%d%d", &x, &y, &r, &w, &c);
            op[i].set(style(s[0]), x, y, c, r, w);
        }
    }
}

inline int get_R (int r, int x, int i, int sign) {
    if (sign == 0)
        return (int)sqrt(1.0 * r * r - 1.0 * (x - i) * (x - i));
    else if (sign == 1)
        return r - abs(x - i);
    else if (sign == 2)
        return r - (i - x);
    return 0;
}

int count (int x, int y, int r, int star, int end, int sign) {
    int ret = 0;
    for (int i = star; i <= end; i++) {
        int R = get_R(r, x, i, sign);

        int mv = max(y - R, 0);
        while (mv = getfar(f[i], mv), abs(mv - y) <= R && mv < M) {
            ret++;
            f[i][mv] = mv+1;
        }
    }
    return ret;
}

int count_rec (int x, int y, int r, int l) {
    int ret = 0;
    for (int i = x; i <= x + r - 1 && i < N; i++) {
        int mv = y;
        while (mv = getfar(f[i], mv), abs(mv - y) < l && mv < M) {
            ret++;
            f[i][mv] = mv+1;
        }
    }
    return ret;
}

void solve () {

    for (int i = Q; i; i--) {
        int& col = cnt[op[i].c];
        if (op[i].sign == 3)
            col += count_rec(op[i].x, op[i].y, op[i].r, op[i].w);
        else
            col += count(op[i].x, op[i].y, op[i].r, op[i].star, op[i].end, op[i].sign);
    }

    for (int i = 1; i <= 9; i++)
        printf("%d%c", cnt[i], i == 9 ? ‘\n‘ : ‘ ‘);
}

int main () {
    while (scanf("%d%d%d", &N, &M, &Q) == 3) {
        init();
        solve();
    }
    return 0;
}

uva 1493 - Draw a Mess(并查集)