UVA 11992 - Fast Matrix Operations
题意:给定一个矩阵,3种操作,在一个矩阵中加入值a,设置值a。查询和
思路:因为最多20列,所以全然能够当作20个线段树来做,然后线段树是区间改动区间查询,利用延迟操作,开两个延迟值一个存放set操作。一个存放add操作
代码:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define lson(x) ((x<<1) + 1) #define rson(x) ((x<<1) + 2) #define INF 0x3f3f3f3f const int N = 1000005; int r, c, m; struct Node { int l, r; int sum, Max, Min, sumv, setv; } node[4 * N]; void pushup(int x) { node[x].sum = node[lson(x)].sum + node[rson(x)].sum; node[x].Max = max(node[lson(x)].Max, node[rson(x)].Max); node[x].Min = min(node[lson(x)].Min, node[rson(x)].Min); } void pushdown(int x) { if (node[x].setv) { node[lson(x)].sumv = node[rson(x)].sumv = 0; node[lson(x)].setv = node[rson(x)].setv = node[x].setv; node[lson(x)].sum = (node[lson(x)].r - node[lson(x)].l + 1) * node[x].setv; node[rson(x)].sum = (node[rson(x)].r - node[rson(x)].l + 1) * node[x].setv; node[lson(x)].Max = node[lson(x)].Min = node[x].setv; node[rson(x)].Max = node[rson(x)].Min = node[x].setv; node[x].setv = 0; } if (node[x].sumv) { node[lson(x)].sumv += node[x].sumv; node[rson(x)].sumv += node[x].sumv; node[lson(x)].sum += (node[lson(x)].r - node[lson(x)].l + 1) * node[x].sumv; node[rson(x)].sum += (node[rson(x)].r - node[rson(x)].l + 1) * node[x].sumv; node[lson(x)].Max += node[x].sumv; node[lson(x)].Min += node[x].sumv; node[rson(x)].Max += node[x].sumv; node[rson(x)].Min += node[x].sumv; node[x].sumv = 0; } } void build(int l, int r, int x) { node[x].l = l; node[x].r = r; if (l == r) { node[x].sum = node[x].Max = node[x].Min = node[x].sumv = node[x].setv = 0; return; } int mid = (l + r) / 2; build(l, mid, lson(x)); build(mid + 1, r, rson(x)); pushup(x); } void add(int l, int r, int v, int x) { if (node[x].l >= l && node[x].r <= r) { node[x].sumv += v; node[x].sum += (node[x].r - node[x].l + 1) * v; node[x].Max += v; node[x].Min += v; return; } pushdown(x); int mid = (node[x].l + node[x].r) / 2; if (l <= mid) add(l, r, v, lson(x)); if (r > mid) add(l, r, v, rson(x)); pushup(x); } void set(int l, int r, int v, int x) { if (node[x].l >= l && node[x].r <= r) { node[x].setv = v; node[x].sum = (node[x].r - node[x].l + 1) * v; node[x].Max = node[x].Min = v; node[x].sumv = 0; return; } pushdown(x); int mid = (node[x].l + node[x].r) / 2; if (l <= mid) set(l, r, v, lson(x)); if (r > mid) set(l, r, v, rson(x)); pushup(x); } Node query(int l, int r, int x) { Node ans; ans.sum = 0; ans.Max = 0; ans.Min = INF; if (node[x].l >= l && node[x].r <= r) { ans.sum = node[x].sum; ans.Max = node[x].Max; ans.Min = node[x].Min; return ans; } pushdown(x); int mid = (node[x].l + node[x].r) / 2; if (l <= mid) { Node tmp = query(l, r, lson(x)); ans.sum += tmp.sum; ans.Max = max(ans.Max, tmp.Max); ans.Min = min(ans.Min, tmp.Min); } if (r > mid) { Node tmp = query(l, r, rson(x)); ans.sum += tmp.sum; ans.Max = max(ans.Max, tmp.Max); ans.Min = min(ans.Min, tmp.Min); } return ans; } int main() { while (~scanf("%d%d%d", &r, &c, &m)) { build(1, r * c, 0); int q, x1, y1, x2, y2, v; while (m--) { scanf("%d", &q); if (q == 3) { scanf("%d%d%d%d", &x1, &y1, &x2, &y2); x1--; x2--; int sum = 0, Max = 0, Min = INF; for (int i = x1; i <= x2; i++) { Node ans = query(i * c + y1, i * c + y2, 0); sum += ans.sum; Max = max(Max, ans.Max); Min = min(Min, ans.Min); } printf("%d %d %d ", sum, Min, Max); } else { scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &v); x1--; x2--; for (int i = x1; i <= x2; i++) { if (q == 1) add(i * c + y1, i * c + y2, v, 0); else set(i * c + y1, i * c + y2, v, 0); } } } } return 0; }
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