一個平面上有 n 個點，每個點有一個對應的權值，其他點的權值都是0，讓你選一個矩形，要求矩形內所有點的權值最大。

枚舉上邊界，對于每一行的點，將他依此加入到線段樹中，同時更新最大值。時間復雜度?

大力區間和并，查詢的時候還要同時查一下跨左右區間的最大值。

#include <bits/stdc++.h> #define lson left,mid,k<<1 #define rson mid+1,right,k<<1|1 #define imid int mid=(que[k].l+que[k].r)/2; #define ll long long #define sc scanf #define pr printf using namespace std; const int MAXN = 2005; struct note {int x;int y;ll val; }in[MAXN]; int t1[MAXN], t2[MAXN]; struct node {int l;int r;ll lmaxn;ll rmaxn;ll maxn;ll sum; }que[MAXN * 4]; int n, qq1, qq2; void up(int k) {que[k].sum = que[k << 1].sum + que[k << 1 | 1].sum;que[k].maxn = max({ que[k << 1].maxn,que[k << 1 | 1].maxn,que[k << 1].rmaxn + que[k << 1 | 1].lmaxn });que[k].lmaxn = max(que[k << 1].lmaxn, que[k << 1].sum + que[k << 1 | 1].lmaxn);que[k].rmaxn = max(que[k << 1 | 1].rmaxn, que[k << 1 | 1].sum + que[k << 1].rmaxn); } void build(int left = 1, int right = n, int k = 1) {que[k].l = left;que[k].r = right;if (left == right){que[k].sum = 0;que[k].maxn = 0;que[k].lmaxn = 0;que[k].rmaxn = 0;return;}imid;build(lson);build(rson);up(k); } void update(int pos, ll val, int k) {if (que[k].l == que[k].r){que[k].sum += val;que[k].maxn += val;que[k].lmaxn += val;que[k].rmaxn += val;return;}imid;if (pos <= mid)update(pos, val, k << 1);if (pos > mid)update(pos, val, k << 1 | 1);up(k); } ll queryl(int left, int right, int k) {if (left == que[k].l && que[k].r == right)return que[k].lmaxn;imid;if (right <= mid)return queryl(left, right, k << 1);else if (left > mid)return queryl(left, right, k << 1 | 1);elsereturn max({ queryl(lson), que[k << 1].maxn + queryl(rson) }); } ll queryr(int left, int right, int k) {if (left == que[k].l && que[k].r == right)return que[k].rmaxn;imid;if (right <= mid)return queryr(left, right, k << 1);else if (left > mid)return queryr(left, right, k << 1 | 1);elsereturn max({ queryr(rson), queryr(lson) + que[k << 1 | 1].maxn }); } ll query(int left, int right, int k) {if (left == que[k].l && que[k].r == right)return que[k].maxn;imid;if (right <= mid)return query(left, right, k << 1);else if (left > mid)return query(left, right, k << 1 | 1);elsereturn max({ query(lson),query(rson),queryr(lson) + queryl(rson) }); } int main() {int T;sc("%d", &T);while (T--){scanf("%d", &n);for (int i = 1; i <= n; i++){scanf("%d%d%lld", &in[i].x, &in[i].y, &in[i].val);t1[i] = in[i].x;t2[i] = in[i].y;}sort(t1 + 1, t1 + 1 + n);qq1 = unique(t1 + 1, t1 + 1 + n) - t1 - 1;for (int i = 1; i <= n; i++)in[i].x = lower_bound(t1 + 1, t1 + 1 + qq1, in[i].x) - t1;sort(t2 + 1, t2 + 1 + n);qq2 = unique(t2 + 1, t2 + 1 + n) - t2 - 1;for (int i = 1; i <= n; i++)in[i].y = lower_bound(t2 + 1, t2 + 1 + qq2, in[i].y) - t2;sort(in + 1, in + 1 + n, [](note q, note w) {if (q.y == w.y)return q.x < w.x;return q.y < w.y;});ll ans = 0;for (int i = 1; i <= qq2; i++){build(1, qq1, 1);for (int j = 1; j <= n; j++){if (in[j].y < i)continue;if (in[j].y != in[j - 1].y){ll res = query(1, qq1, 1);ans = max(res, ans);}update(in[j].x, in[j].val, 1);}ll res = query(1, qq1, 1);ans = max(res, ans);}pr("%lld\n", ans);} } /* 1 9 1 1 1 1 2 1 1 3 1 2 1 1 2 2 -9 2 3 1 3 1 1 3 2 1 3 3 1 */

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