UVA11992 - Fast Matrix Operations(线段树区间改动)
题目大意：给你个r*c的矩阵，初始化为0。
然后给你三种操作：
1 x1, y1, x2, y2, v 把由x1，y1, x2, y2构成的子矩阵里的每一个元素都加上v。
2 x1, y1, x2, y2, v 把这个子矩阵的每一个元素都改动为v。
3 x1, y1, x2, y2 查询这个子矩阵的元素之和，和这些元素的最大值和最小值。
解题思路：由于矩阵的行最多20行，所以能够将这个矩阵的元素构建一棵线段树，每一个节点都有三个附加信息：sum，Max_num, Min_num.然后改动查询的时候就每行每行的来做。注意：每次set的时候都要将这个节点的add清理掉。由于add已经是不须要的了。
然后每次的pushdown都要先处理set再处理add。pushdown处理的是这个节点的子节点。然后通过pushup就能够更新这个节点的附加信息。
代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1) + 1)
const int N = 1e6 + 5;
const int INF = 0x3f3f3f3f;
struct Item {
int sum, Max_num, Min_num;
Item (int sum = 0, int Max_num = -INF, int Min_num = INF) {
this->set(sum, Max_num, Min_num);
}
void set (int sum, int Max_num, int Min_num) {
this->sum = sum;
this->Max_num = Max_num;
this->Min_num = Min_num;
}
};
struct Node {
Item v;
int l, r;
int addv, setv;
void set (int l, int r, int addv = 0, int setv = -1) {
this->l = l;
this->r = r;
this->addv = addv;
this->setv = setv;
}
}node[4 * N];
Item get_item (Item a, Item b) {
return Item(a.sum + b.sum, max(a.Max_num, b.Max_num), min(a.Min_num, b.Min_num));
}
void Node_add(int u, int addv) {
node[u].addv += addv;
node[u].v.sum += addv * (node[u].r - node[u].l + 1);
node[u].v.Max_num += addv;
node[u].v.Min_num += addv;
}
void Node_set(int u, int setv) {
node[u].setv = setv;
node[u].addv = 0;
node[u].v.sum = setv * (node[u].r - node[u].l + 1);
node[u].v.Max_num = node[u].v.Min_num = setv;
}
void pushup (int u) {
node[u].set(node[lson(u)].l, node[rson(u)].r);
node[u].v = get_item(node[lson(u)].v, node[rson(u)].v);
}
void pushdown(int u) {
if (node[u].setv >= 0) {
Node_set(lson(u), node[u].setv);
Node_set(rson(u), node[u].setv);
node[u].setv = -1;
}
if (node[u].addv) {
Node_add(lson(u), node[u].addv);
Node_add(rson(u), node[u].addv);
node[u].addv = 0;
}
}
void Build (int u, int l, int r) {
if (l == r) {
node[u].set(l, r);
node[u].v.set(0, 0, 0);
} else {
int m = (l + r) / 2;
Build(lson(u), l, m);
Build(rson(u), m + 1, r);
pushup(u);
}
}
void Add (int u, int l, int r, int addv) {
if (node[u].l >= l && node[u].r <= r)
Node_add(u, addv);
else {
pushdown(u);
int m = (node[u].l + node[u].r) / 2;
if (l <= m)
Add (lson(u), l, r, addv);
if (r > m)
Add (rson(u), l, r, addv);
pushup(u);
}
}
void Set (int u, int l, int r, int setv) {
if (node[u].l >= l && node[u].r <= r)
Node_set(u, setv);
else {
pushdown(u);
int m = (node[u].l + node[u].r) / 2;
if (l <= m)
Set (lson(u), l, r, setv);
if (r > m)
Set (rson(u), l, r, setv);
pushup(u);
}
}
Item Query (int u, int l, int r) {
if (node[u].l >= l && node[u].r <= r)
return node[u].v;
else {
Item ans;
pushdown(u);
int m = (node[u].l + node[u].r) / 2;
if (l <= m)
ans = get_item (ans, Query(lson(u), l, r));
if (r > m)
ans = get_item (ans, Query(rson(u), l, r));
pushup(u);
return ans;
}
}
int main () {
int r, c, m;
int type;
int x1, y1, x2, y2, v;
while (scanf ("%d%d%d", &r, &c, &m) != EOF) {
Build(1, 1, r * c);
while (m--) {
scanf ("%d", &type);
if (type == 1) {
scanf ("%d%d%d%d%d", &x1, &y1, &x2, &y2, &v);
for (int i = x1 - 1; i < x2; i++)
Add(1, i * c + y1, i * c + y2, v);
} else if (type == 2) {
scanf ("%d%d%d%d%d", &x1, &y1, &x2, &y2, &v);
for (int i = x1 - 1; i < x2; i++)
Set(1, i * c + y1, i * c + y2, v);
} else {
scanf ("%d%d%d%d", &x1, &y1, &x2, &y2);
Item ans;
for (int i = x1 - 1; i < x2; i++)
ans = get_item(ans, Query(1, i * c + y1, i * c + y2));
printf ("%d %d %d\n", ans.sum, ans.Min_num, ans.Max_num);
}
}
}
return 0;
}版权声明：本文博主原创文章，博客，未经同意不得转载。