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区间合并。
需要记录三个东西，sum[rt] 表示这个区间最长连续空集的长度，lsum[rt]=表示从左边开始最长连续空集的长度，rum[rt]表示从右边最长连续空集的长度。
#include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1const int maxn = 55555;int color[maxn << 2];int sum[maxn << 2], lsum[maxn << 2], rsum[maxn << 2];void down(int rt, int m){ if (~color[rt]){ color[rt << 1] = color[rt << 1 | 1] = color[rt]; sum[rt << 1] = lsum[rt << 1] = rsum[rt << 1] = color[rt] ? 0 : (m - (m >> 1)); sum[rt << 1 | 1] = lsum[rt << 1 | 1] = rsum[rt << 1 | 1] = color[rt] ? 0 : (m >> 1); color[rt] = -1; }}void up(int rt, int m){ lsum[rt] = lsum[rt << 1]; rsum[rt] = rsum[rt << 1 | 1]; if (lsum[rt] == (m - (m >> 1))) lsum[rt] += lsum[rt << 1 | 1]; if (rsum[rt] == (m >> 1)) rsum[rt] += rsum[rt << 1]; sum[rt] = max(rsum[rt << 1] + lsum[rt << 1 | 1], max(sum[rt << 1], sum[rt << 1 | 1]));//这里就是很神，不知道怎么想到的，用这三个量就可以记录更新区间最长连续0。sum的大小只和左右儿子的sum，和横跨左右儿子的那一部分有关。}void build(int l, int r, int rt){ color[rt] = -1; sum[rt] = lsum[rt] = rsum[rt] = r - l + 1; if (l == r) return; int mid = (l + r) >> 1; build(lson); build(rson);}void update(int L, int R, int add, int l, int r, int rt){ if (L <= l&&r <= R){ sum[rt] = lsum[rt] = rsum[rt] = add ? 0 : r - l + 1; color[rt] = add; return; } down(rt, r - l + 1); int mid = (l + r) >> 1; if (L <= mid) update(L, R, add, lson); if (R>mid) update(L, R, add, rson); up(rt, r - l + 1);}int ask(int key, int l, int r, int rt){ if (l == r) return l; int mid = (l + r) >> 1; down(rt, r - l + 1); if (sum[rt << 1] >= key) return ask(key, lson); else if (rsum[rt << 1] + lsum[rt << 1 | 1] >= key) return mid - rsum[rt << 1] + 1; return ask(key, rson);}int main(){ int n, m; int a, b, c; while (cin >> n >> m){ build(1, n, 1); for (int i = 0; i < m; i++){ scanf("%d", &a); if (a == 1){ scanf("%d", &b); if (b>sum[1]){ printf("0\n"); continue; } int pos = ask(b, 1, n, 1); printf("%d\n", pos); update(pos, pos + b - 1, 1, 1, n, 1); } else{ scanf("%d%d", &b, &c); update(b, b + c - 1, 0, 1, n, 1); } } } return 0;}
转载于:https://www.cnblogs.com/yigexigua/p/3928287.html