并查集(涂色问题) HDOJ 4056 Draw a Mess

题目传送门

题意:给出一个200 * 50000的像素点矩阵，执行50000次操作，每次把一个矩形/圆形/菱形/三角形内的像素点涂成指定颜色，问最后每种颜色的数量。

分析:乍一看,很像用线段树成段更新写,虽然复杂度有点大,但是也想不到其他的方法.这题可以巧妙地运用并查集来涂色.离线,从最后一个倒过来涂色,保证能涂上去的一定不会被覆盖,每次扫描一行,将一条线上的点都合并到右端点.

#include <bits/stdc++.h>
using namespace std;

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
typedef long long LL;
const int N = 5e4 + 2;
const int INF = 0x3f3f3f3f;
struct DSU  {
    int rt[N];
    void clear(int n) {
        fill (rt, rt+n, -1);
    }
    int Find(int x) {
        return rt[x] == -1 ? x : rt[x] = Find (rt[x]);
    }
    void Union(int x, int y)    {
        rt[x] = y;
    }
}dsu;
struct Point    {
    char str[20];
    int xc, yc, a, b, c;
}p[N];
bool vis[N];
int ans[10];
int n, m, q;

int squ(int x)	{
	return x * x;
}

int cal(int x, int y)	{
	x = abs (x);	y = abs (y);
	return squ (x) + squ (y);
}

int main(void)	{       //好题
	while (scanf ("%d%d%d", &n, &m, &q) == 3)	{
		memset (ans, 0, sizeof (ans));
        for (int i=0; i<q; ++i) {
            scanf ("%s%d%d", &p[i].str, &p[i].xc, &p[i].yc);
            if (p[i].str[0] == 'R') scanf ("%d%d%d", &p[i].a, &p[i].b, &p[i].c);
            else    scanf ("%d%d", &p[i].a, &p[i].c);
        }
        for (int i=0; i<n; ++i) {
            dsu.clear (m);   memset (vis, false, sizeof (vis));
            for (int j=q-1; j>=0; --j) {
                int left = 0, right = 0;
                if (p[j].str[0] == 'C') {
                    if (squ (i - p[j].xc) > squ (p[j].a))    continue;
                    int tmp = (int) sqrt (1.0 * (squ (p[j].a) - squ (i - p[j].xc)));
                    left = p[j].yc - tmp;   right = p[j].yc + tmp;
                }
                else if (p[j].str[0] == 'D') {
                    if (i < p[j].xc - p[j].a || i > p[j].xc + p[j].a)   continue;
                    left = p[j].yc - p[j].a + abs (i - p[j].xc);    right = p[j].yc + p[j].a - abs (i - p[j].xc);
                }
                else if (p[j].str[0] == 'R') {
                    if (i < p[j].xc || i > p[j].xc + p[j].a - 1) continue;
                    left = p[j].yc; right = p[j].yc + p[j].b - 1;
                }
                else if (p[j].str[0] == 'T') {
                    if (i < p[j].xc || i > p[j].xc + (p[j].a + 1) / 2 - 1) continue;
                    left = p[j].yc - p[j].a / 2 + (i - p[j].xc); right = p[j].yc + p[j].a / 2 - (i - p[j].xc);
                }
                if (left < 0)   left = 0;
                if (right >= m)  right = m - 1;
                int fa, fb = dsu.Find (right);
                for (int k=left; k<=right; k=fa+1) {        //paint a line
                    fa = dsu.Find (k);
                    if (!vis[fa])   {
                        ans[p[j].c]++;  vis[fa] = true;
                    }
                    if (fa != fb)   dsu.Union (fa, fb);
                }
            }
        }
		for (int i=1; i<=9; ++i)	{
			printf ("%d%c", ans[i], i == 9 ? '\n' : ' ');
		}
	}

	return 0;
}