UVA 11992 - Fast Matrix Operations

题目链接

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 0x3f3f3f3fconst 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\n", 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;
}