uva 1493(并查集)

题意：一个画板上可以画4种图形，画板被分成n*m个网格，给出了每种图形的坐标大小和颜色，画了q次后，问画板上每种颜色占多少格子。
题解：因为只要最后被覆盖的颜色，所以倒着操作，这样已经画过的地方就不再计算，未画过的就并到已画的集合内，方便统计数量。

#include <stdio.h>#include <algorithm>#include <string.h>#include <cmath>using namespace std;const int N = 50005;int n, m, q, color[10], pa[205][N];char str[20];struct S {    int xc, yc, r, l, w, c, flag;}s[N];int get_parent(int *pa, int x) {    return pa[x] == x ? x : pa[x] = get_parent(pa, pa[x]);}void solve(int l, int r, int *pa, int c) {    l = get_parent(pa, l);    while (l <= r) {        color[c]++;        pa[l] = get_parent(pa, l + 1);        l = pa[l];    }}void count_d(int xc, int yc, int r, int c) {    for (int x = max(0, xc - r); x <= min(n - 1, xc + r); x++) {        int rr = r - abs(x - xc);        int l1 = max(0, yc - rr), r1 = min(m - 1, yc + rr);        solve(l1, r1, pa[x], c);    }}void count_c(int xc, int yc, int r, int c) {    for (int x = max(0, xc - r); x <= min(n - 1, xc + r); x++) {        int rr = (int)(sqrt(r * r - (x - xc) * (x - xc)));        int l1 = max(0, yc - rr), r1 = min(m - 1, yc + rr);        solve(l1, r1, pa[x], c);    }}void count_t(int xc, int yc, int w, int c) {    for (int x = max(0, xc); x <= min(n - 1, xc + (w + 1) / 2 - 1); x++) {        int ww = ((w + 1) / 2 - 1) - (x - xc);        int l1 = max(0, yc - ww), r1 = min(m - 1, yc + ww);        solve(l1, r1, pa[x], c);    }}void count_r(int xc, int yc, int l, int w, int c) {    for (int x = max(0, xc); x <= min(n - 1, xc + l - 1); x++) {        int l1 = max(0, yc), r1 = min(m - 1, yc + w - 1);        solve(l1, r1, pa[x], c);    }}int main() {    while (scanf("%d%d%d", &n, &m, &q) == 3) {        for (int i = 0; i < n; i++)                 for (int j = 0; j <= m; j++)                pa[i][j] = j;        memset(color, 0, sizeof(color));        for (int i = 0; i < q; i++) {            scanf("%s", str);            if (str[0] == 'R') {                scanf("%d%d%d%d%d", &s[i].xc, &s[i].yc, &s[i].l, &s[i].w, &s[i].c);                s[i].flag = 3;            }            else if (str[0] == 'T') {                scanf("%d%d%d%d", &s[i].xc, &s[i].yc, &s[i].r, &s[i].c);                s[i].flag = 2;            }            else if (str[0] == 'C') {                scanf("%d%d%d%d", &s[i].xc, &s[i].yc, &s[i].r, &s[i].c);                s[i].flag = 1;            }            else {                scanf("%d%d%d%d", &s[i].xc, &s[i].yc, &s[i].r, &s[i].c);                s[i].flag = 0;            }        }        for (int i = q - 1; i >= 0; i--) {            if (s[i].flag == 0)                count_d(s[i].xc, s[i].yc, s[i].r, s[i].c);            else if (s[i].flag == 1)                count_c(s[i].xc, s[i].yc, s[i].r, s[i].c);            else if (s[i].flag == 2)                count_t(s[i].xc, s[i].yc, s[i].r, s[i].c);            else if (s[i].flag == 3)                count_r(s[i].xc, s[i].yc, s[i].l, s[i].w, s[i].c);        }        printf("%d", color[1]);        for (int i = 2; i < 10; i++)            printf(" %d", color[i]);        printf("\n");    }    return 0;}