UVA 1493 - Draw a Mess(并查集)

UVA 1493 - Draw a Mess

题目链接

题意：在一个n*m平面上，有4种操作，对应把相应区域颜色涂成v(1<= v <= 9)，现在经过q次操作以后，要求9种颜色各有多少个

思路：并查集，由于颜色涂上去会覆盖，这样我们就可以反向执行操作，这样保证每次操作如果之前有颜色就不能涂，如果没有就可以涂，然后一共有200行，每行都利用并查集压缩路径，查找下一个能涂色的位置即可

题目中说三角形边一定是奇数，可居然有偶数的。。。被这个坑了

代码：

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