UVA_11983
这个题目可以转化成求被覆盖K次的矩形面积的并，只要将x2和y2都+1即可，相当于我们把每个单位正方形看成一个点。
#include#include #include #define MAXK 11#define MAXD 60010int N, K, M, len[4 * MAXD][MAXK], cnt[4 * MAXD], ty[MAXD];struct Seg{ int x, y1, y2, col;}seg[MAXD];int cmpint(const void *_p, const void *_q){ int *p = (int *)_p, *q = (int *)_q; return *p < *q ? -1 : 1;}int cmps(const void *_p, const void *_q){ Seg *p = (Seg *)_p, *q = (Seg *)_q; return p->x < q->x ? -1 : 1;}void build(int cur, int x, int y){ int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1; memset(len[cur], 0, sizeof(len[cur])); len[cur][0] = ty[y + 1] - ty[x]; cnt[cur] = 0; if(x == y) return ; build(ls, x, mid); build(rs, mid + 1, y);}void init(){ int i, j, k, x1, y1, x2, y2; scanf("%d%d", &N, &K); for(i = 0; i < N; i ++) { j = i << 1, k = (i << 1) | 1; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); ++ x2, ++ y2; seg[j].x = x1, seg[k].x = x2; seg[j].y1 = seg[k].y1 = y1, seg[j].y2 = seg[k].y2 = y2; seg[j].col = 1, seg[k].col = -1; ty[j] = y1, ty[k] = y2; } qsort(ty, N << 1, sizeof(ty[0]), cmpint); M = -1; for(i = 0; i < (N << 1); i ++) if(i == 0 || ty[i] != ty[i - 1]) ty[++ M] = ty[i]; build(1, 0, M - 1);}int BS(int x){ int mid, min = 0, max = M + 1; for(;;) { mid = (min + max) >> 1; if(mid == min) break; if(ty[mid] <= x) min = mid; else max = mid; } return mid;}void update(int cur, int x, int y){ int ls = cur << 1, rs = (cur << 1) | 1; memset(len[cur], 0, sizeof(len[cur])); if(cnt[cur] >= K) len[cur][K] = ty[y + 1] - ty[x]; else if(x == y) len[cur][cnt[cur]] = ty[y + 1] - ty[x]; else { int i; for(i = cnt[cur]; i <= K; i ++) len[cur][i] += len[ls][i - cnt[cur]] + len[rs][i - cnt[cur]]; for(i = K - cnt[cur] + 1; i <= K; i ++) len[cur][K] += len[ls][i] + len[rs][i]; }}void refresh(int cur, int x, int y, int s, int t, int c){ int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1; if(x >= s && y <= t) { cnt[cur] += c; update(cur, x, y); return ; } if(mid >= s) refresh(ls, x, mid, s, t, c); if(mid + 1 <= t) refresh(rs, mid + 1, y, s, t, c); update(cur, x, y);}void solve(){ int i, j, k; long long int ans = 0; qsort(seg, N << 1, sizeof(seg[0]), cmps); seg[N << 1].x = seg[(N << 1) - 1].x; for(i = 0; i < (N << 1); i ++) { j = BS(seg[i].y1), k = BS(seg[i].y2); refresh(1, 0, M - 1, j, k - 1, seg[i].col); ans += (long long int)len[1][K] * (seg[i + 1].x - seg[i].x); } printf("%lld\n", ans);}int main(){ int t, tt; scanf("%d", &t); for(tt = 0; tt < t; tt ++) { init(); printf("Case %d: ", tt + 1); solve(); } return 0;}