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题意:训练指南P207
分析:因为矩阵不超过20行,所以可以建20条线段的线段树,支持两个区间更新以及区间查询.
#include <bits/stdc++.h> using namespace std; #define lson l, mid, o << 1 #define rson mid + 1, r, o << 1 | 1 typedef long long ll; const int INF = 0x3f3f3f3f; const int N = 20 + 5; const int M = 5e4 + 5; struct Segment_Tree { struct Node { int sum, mx, mn, add, setv; }node[M<<2]; void _max(int &a, int b) { if (a < b) a = b; } void _min(int &a, int b) { if (a > b) a = b; } void push_up(int o) { node[o].sum = node[o<<1].sum + node[o<<1|1].sum; node[o].mx = max (node[o<<1].mx, node[o<<1|1].mx); node[o].mn = min (node[o<<1].mn, node[o<<1|1].mn); } void push_down(int l, int r, int o) { int len = r - l + 1; if (node[o].setv != -1 && l < r) { //先set,然后再add node[o<<1].setv = node[o<<1|1].setv = node[o].setv; node[o<<1].add = node[o<<1|1].add = 0; //很重要的地方,有set的清除add,意思是set前的add都不要了 node[o<<1].sum = node[o].setv * (len - (len >> 1)); node[o<<1].mx = node[o<<1].mn = node[o].setv; node[o<<1|1].sum = node[o].setv * (len >> 1); node[o<<1|1].mx = node[o<<1|1].mn = node[o].setv; node[o].setv = -1; } if (node[o].add != 0 && l < r) { node[o<<1].add += node[o].add; node[o<<1|1].add += node[o].add; node[o<<1].sum += node[o].add * (len - (len >> 1)); node[o<<1].mx += node[o].add; node[o<<1].mn += node[o].add; node[o<<1|1].sum += node[o].add * (len >> 1); node[o<<1|1].mx += node[o].add; node[o<<1|1].mn += node[o].add; node[o].add = 0; } } void build(int l, int r, int o) { node[o].add = 0; node[o].setv = -1; if (l == r) { node[o].sum = 0; node[o].mx = 0; node[o].mn = 0; return ; } int mid = l + r >> 1; build (lson); build (rson); push_up (o); } void updata(int ql, int qr, int op, int c, int l, int r, int o) { if (ql <= l && r <= qr) { if (op == 1) { node[o].sum += c * (r - l + 1); node[o].mx += c; node[o].mn += c; node[o].add += c; return ; } else if (op == 2) { node[o].sum = c * (r - l + 1); node[o].mx = node[o].mn = c; node[o].add = 0; node[o].setv = c; return ; } } push_down (l, r, o); int mid = l + r >> 1; if (ql <= mid) updata (ql, qr, op, c, lson); if (qr > mid) updata (ql, qr, op, c, rson); push_up (o); } void query(int ql, int qr, int &_sum, int &_mn, int &_mx, int l, int r, int o) { if (ql <= l && r <= qr) { _sum += node[o].sum; _max (_mx, node[o].mx); _min (_mn, node[o].mn); return ; } push_down (l, r, o); int mid = l + r >> 1; if (ql <= mid) query (ql, qr, _sum, _mn, _mx, lson); if (qr > mid) query (ql, qr, _sum, _mn, _mx, rson); } }st[N]; int n, m, q; int main(void) { while (scanf ("%d%d%d", &n, &m, &q) == 3) { for (int i=1; i<=n; ++i) { st[i].build (1, m, 1); } int x1, y1, x2, y2, v, op; while (q--) { scanf ("%d%d%d%d%d", &op, &x1, &y1, &x2, &y2); if (op <= 2) { scanf ("%d", &v); for (int i=x1; i<=x2; ++i) { st[i].updata (y1, y2, op, v, 1, m, 1); } } else { int sum = 0, mx = 0, mn = INF; for (int i=x1; i<=x2; ++i) { st[i].query (y1, y2, sum, mn, mx, 1, m, 1); } printf ("%d %d %d\n", sum, mn, mx); } } } return 0; }
线段树(多维+双成段更新) UVA 11992 Fast Matrix Operations
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原文地址:http://www.cnblogs.com/Running-Time/p/5048016.html