线段树区间更新——UVA 11992

Fast Matrix Operations

Time Limit:5000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status

Description

Problem F

Fast Matrix Operations

There is a matrix containing at most 10⁶ elements divided into r rows and c columns. Each element has a location (x,y) where 1<=x<=r,1<=y<=c. Initially, all the elements are zero. You need to handle four kinds of operations:

1 x1 y1 x2 y2 v

Increment each element (x,y) in submatrix (x1,y1,x2,y2) by v (v>0)

2 x1 y1 x2 y2 v

Set each element (x,y) in submatrix (x1,y1,x2,y2) to v

3 x1 y1 x2 y2

Output the summation, min value and max value of submatrix (x1,y1,x2,y2)

In the above descriptions, submatrix (x1,y1,x2,y2) means all the elements (x,y) satisfying x1<=x<=x2 and y1<=x<=y2. It is guaranteed that 1<=x1<=x2<=r, 1<=y1<=y2<=c. After any operation, the sum of all the elements in the matrix does not exceed 10⁹.

Input

There are several test cases. The first line of each case contains three positive integers r, c, m, where m (1<=m<=20,000) is the number of operations. Each of the next m lines contains a query. There will be at most twenty rows in the matrix. The input is terminated by end-of-file (EOF). The size of input file does not exceed 500KB.

Output

For each type-3 query, print the summation, min and max.

Sample Input

4 4 81 1 2 4 4 53 2 1 4 41 1 1 3 4 23 1 2 4 43 1 1 3 42 2 1 4 4 23 1 2 4 41 1 1 4 3 3

Output for the Sample Input

45 0 578 5 769 2 739 2 7

题意：对一个r*c的矩阵，元素个数不会超过10^6，有3个操作，任何运算的和不会超过10^9。

1  x1, y1, x2, y2, v  (x1 <= x <= x2, y1 <= y <= y2) 表示把此子矩阵里的元素全加上v。

2  x1, y1, x2, y2, v  (x1 <= x <= x2, y1 <= y <= y2) 表示把此子矩阵里的元素全变成v。

3  x1, y1, x2, y2,     (x1 <= x <= x2, y1 <= y <= y2) 表示求此子矩阵的最大元素，最小元素，所有元素和。

PS：题目描述有点不具体，虽然影响不大。。。（没有说明操作2的v值范围，还有后面的y1 <= x <= y2）

思路： 把多维展开成一维，就相当于每次询问就要查询 x2 - x1 + 1 次（一行一行地更新）。总区间为 r*c ，每次查询区间为left = y1 + c*(x-1);    right = y2 + c*(x - 1);  这样就变成普通的区间覆盖和增加问题。只需要注意一点，就是再遇到覆盖时就要把增加标记置0。这样，在往下更新时就先判断覆盖，再判断增加。

#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX(x,y) ((x)>(y)?(x):(y))#define MIN(x,y) ((x)<(y)?(x):(y))#define MS(x,y) memset(x, y, sizeof(x))const int MAXN=1000100;const int INF=1000000005;int sum[MAXN<<2];int min[MAXN<<2];int max[MAXN<<2];int add[MAXN<<2];int cover[MAXN<<2];void down(int root, int len){int ll = root<<1;int rr = root<<1|1;int cov = cover[root];int ad = add[root];if(cov){//有覆盖就要向下更新cover[ll] = cover[rr] = cov;sum[ll] = cov * (len - (len>>1));sum[rr] = cov * (len>>1);max[ll] = max[rr] = cov;min[ll] = min[rr] = cov;add[ll] = add[rr] = 0;//把增加置空cover[root] = 0;}if(ad){//有增加就说明最后的操作不是覆盖add[ll] += ad;add[rr] += ad;sum[ll] += ad * (len - (len>>1));sum[rr] += ad * (len>>1);max[ll] += ad;max[rr] += ad;min[ll] += ad;min[rr] += ad;add[root] = 0;}}void up(int root){int ll = root<<1;int rr = root<<1|1;sum[root] = sum[ll] + sum[rr];max[root] = MAX(max[ll], max[rr]);min[root] = MIN(min[ll], min[rr]);}void IS(int root, int left, int right, int l, int r, int v, int flag){if(l == left && right == r){if(1 == flag){add[root] += v;sum[root] += v * (right - left + 1);max[root] += v;min[root] += v;}else{cover[root] = v;add[root] = 0;//把增加置空sum[root] = v * (right - left + 1);max[root] = v;min[root] = v;}return;}down(root, right - left + 1);int mid = (left + right)>>1;if(r <= mid) IS(root<<1, left, mid, l, r, v, flag);else if(l > mid) IS(root<<1|1, mid + 1, right, l, r, v, flag);else{IS(root<<1, left, mid, l, mid, v, flag);IS(root<<1|1, mid + 1, right, mid + 1, r, v, flag);}up(root);}typedef struct{int sum, min, max;}Node;Node query(int root, int left, int right, int l, int r){Node N,NL,NR;if(l == left && right == r){N.sum = sum[root];N.max = max[root];N.min = min[root];return N;}down(root, right - left + 1);int mid = (left + right)>>1;if(r <= mid) return N = query(root<<1, left, mid, l, r);else if(l > mid) return N = query(root<<1|1, mid+1, right, l, r);else{NL = query(root<<1, left, mid, l, mid);NR = query(root<<1|1, mid+1, right, mid+1, r);N.sum = NL.sum + NR.sum;N.max = MAX(NL.max, NR.max);N.min = MIN(NL.min, NR.min);return N;}up(root);}int main(){//freopen("in.txt","r",stdin);int r,c,m,n;while(~scanf("%d%d%d", &r,&c,&m)){MS(add, 0);MS(cover, 0);MS(sum, 0);MS(max, 0);MS(min, 0);n = r*c;int o,x1,y1,x2,y2,v;int i,beg,end,Sum,Min,Max;Node node;while(m--){scanf("%d", &o);if(1 == o){scanf("%d%d%d%d%d", &x1,&y1,&x2,&y2,&v);for(i=x1; i<=x2; i++){beg = y1 + c * (i - 1);end = y2 + c * (i - 1);IS(1, 1, n, beg, end, v, 1);}}else if(2 == o){scanf("%d%d%d%d%d", &x1,&y1,&x2,&y2,&v);for(i=x1; i<=x2; i++){beg = y1 + c * (i - 1);end = y2 + c * (i - 1);IS(1, 1, n, beg, end, v, 0);}}else{scanf("%d%d%d%d", &x1,&y1,&x2,&y2);Sum = 0;Max = 0;Min = INF;for(i=x1; i<=x2; i++){beg = y1 + c * (i - 1);end = y2 + c * (i - 1);node = query(1, 1, n, beg, end);Sum += node.sum;Max = MAX(Max, node.max);Min = MIN(Min, node.min);}printf("%d %d %d\n", Sum, Min, Max);}}}return 0;}