hdu 5454 Excited Database（线段树）

题目链接：hdu 5454 Excited Database

解题思路

维护两科线段树，一棵i+j，一棵i-j。因为对角线上的值一定都是相同的，只是查询时的个数不一样，但是根据区间范围可以计算出来每条对角线的个数。每个树维护单一节点权值和s，从右边递增的和r，从左边递增的和l。查询的时候将矩形分成3份。

代码

#include <cstdio>
#include <cstring>
#include <algorithm>using namespace std;
const int maxn = 400005;
typedef long long ll;
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)struct SegTree {int lc[maxn<<2], rc[maxn<<2];ll S[maxn<<2], L[maxn<<2], R[maxn<<2], T[maxn<<2];int length(int u) {return rc[u] - lc[u] + 1;}void maintain(int u, ll a) {T[u] += a;S[u] += a * length(u);ll add = a * (length(u) + 1) * length(u) / 2;L[u] += add;R[u] += add;}void pushup(int u) {S[u] = S[lson(u)] + S[rson(u)];L[u] = L[lson(u)] + L[rson(u)] + S[rson(u)] * length(lson(u));R[u] = R[rson(u)] + R[lson(u)] + S[lson(u)] * length(rson(u));}void pushdown(int u) {if (T[u]) {maintain(lson(u), T[u]);maintain(rson(u), T[u]);T[u] = 0;}}void build (int u, int l, int r) {lc[u] = l, rc[u] = r;S[u] = L[u] = R[u] = T[u] = 0;if (l == r) return;int mid = (l+r)>>1;build (lson(u), l, mid);build (rson(u), mid+1, r);pushup(u);}void modify(int u, int l, int r, int v) {if (l <= lc[u] && rc[u] <= r) {maintain(u, v);return;}pushdown(u);int mid = (lc[u] + rc[u]) >> 1;if (l <= mid) modify(lson(u), l, r, v);if (r > mid) modify(rson(u), l, r, v);pushup(u);}ll query(int u, int l, int r) {if (l <= lc[u] && rc[u] <= r)return S[u];pushdown(u);int mid = (lc[u] + rc[u]) >> 1;ll ret = 0;if (l <= mid) ret += query(lson(u), l, r);if (r > mid) ret += query(rson(u), l, r);pushup(u);return ret;}ll queryLeft(int u, int l, int r) {if (l <= lc[u] && rc[u] <= r)return L[u] + S[u] * (lc[u] - l);pushdown(u);int mid = (lc[u] + rc[u]) >> 1;ll ret = 0;if (l <= mid) ret += queryLeft(lson(u), l, r);if (r > mid) ret += queryLeft(rson(u), l, r);pushup(u);return ret;}ll queryRight(int u, int l, int r) {if (l <= lc[u] && rc[u] <= r)return R[u] + S[u] * (r - rc[u]);pushdown(u);int mid = (lc[u] + rc[u]) >> 1;ll ret = 0;if (l <= mid) ret += queryRight(lson(u), l, r);if (r > mid) ret += queryRight(rson(u), l, r);pushup(u);return ret;}
}A, D;int N, M;int main () {int cas;scanf("%d", &cas);for (int kcas = 1; kcas <= cas; kcas++) {scanf("%d%d", &N, &M);A.build(1, 2, 2 * N);D.build(1, 1, 2 * N - 1);
//      A.build(1, 1, 2 * N);
//      D.build(1, 1, 2 * N);printf("Case #%d:\n", kcas);int t, l, r, x1, y1, x2, y2;while (M--) {scanf("%d", &t);if (t == 1) {scanf("%d%d", &l, &r);A.modify(1, l, r, 1);} else if (t == 2) {scanf("%d%d", &l, &r);D.modify(1, l + N, r + N, 1);} else {scanf("%d%d%d%d", &x1, &x2, &y1, &y2);ll ans = 0, t = min(x2-x1, y2-y1) + 1;;l = x2 + y1, r = x1 + y2;if (l > r) swap(l, r);ans += A.query(1, l, r) * t;ans += A.queryLeft(1, x1+y1, l-1);ans += A.queryRight(1, r+1, x2+y2);l = x1-y1+N, r = x2-y2+N;if (l > r) swap(l, r);ans += D.query(1, l, r) * t;ans += D.queryLeft(1, x1-y2+N, l-1);ans += D.queryRight(1, r+1, x2-y1+N);printf("%lld\n", ans);}}}return 0;
}

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hdu 5454 Excited Database（线段树）

解题思路

代码

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