UVa 11992 - Fast Matrix Operations 成段更新,求最值与和
来源:互联网 发布:java小项目开发案例 编辑:程序博客网 时间:2024/06/14 18:24
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
Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yeji Shen, Dun Liang
Note: Please make sure to test your program with the gift I/O files before submitting!
-----------------------
模板
-----------------------
#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#define N 555555using namespace std;const int OO=1e9;int num[30][N];int _min,_max,_sum;struct Tree{ int l; int r; int max; int min; int sum; int add; int set;}big_tree[21][N*4];void push_up(int root,Tree tree[]){ tree[root].max=max(tree[root<<1].max,tree[root<<1|1].max); tree[root].min=min(tree[root<<1].min,tree[root<<1|1].min); tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum;}void push_down(int root,Tree tree[]){ if (tree[root].set!=-1) { if (tree[root].l!=tree[root].r) { //传递懒惰标记 tree[root<<1].add=tree[root<<1|1].add=0; tree[root<<1].set=tree[root<<1|1].set=tree[root].set; //最更新大值 tree[root<<1].max=tree[root<<1|1].max=tree[root].set; //更新最小值 tree[root<<1].min=tree[root<<1|1].min=tree[root].set; //更新区间和 tree[root<<1].sum=(tree[root<<1].r-tree[root<<1].l+1)*tree[root].set; tree[root<<1|1].sum=(tree[root<<1|1].r-tree[root<<1|1].l+1)*tree[root].set; } tree[root].set=-1; } if (tree[root].add>0) { if (tree[root].l!=tree[root].r) { //传递懒惰标记 tree[root<<1].add+=tree[root].add; tree[root<<1|1].add+=tree[root].add; //更新最大值 tree[root<<1].max+=tree[root].add; tree[root<<1|1].max+=tree[root].add; //更新最小值 tree[root<<1].min+=tree[root].add; tree[root<<1|1].min+=tree[root].add; //更新区间和 tree[root<<1].sum+=(tree[root<<1].r-tree[root<<1].l+1)*tree[root].add; tree[root<<1|1].sum+=(tree[root<<1|1].r-tree[root<<1|1].l+1)*tree[root].add; } tree[root].add=0; }}void build(int root,int l,int r,Tree tree[]){ tree[root].l=l; tree[root].r=r; if(tree[root].l==tree[root].r) { tree[root].max=0; tree[root].min=0; tree[root].sum=0; tree[root].add=0; tree[root].set=-1; return; } int mid=(l+r)/2; build(root<<1,l,mid,tree); build(root<<1|1,mid+1,r,tree); push_up(root,tree);}void update_add(int root,int L,int R,int val,Tree tree[]){ if(L<=tree[root].l&&R>=tree[root].r) { tree[root].add+=val; tree[root].max+=val; tree[root].min+=val; tree[root].sum+=(tree[root].r-tree[root].l+1)*val; return; } push_down(root,tree); int mid=(tree[root].l+tree[root].r)/2; if(L<=mid) update_add(root<<1,L,R,val,tree); if (R>mid) update_add(root<<1|1,L,R,val,tree); push_up(root,tree);}void update_set(int root,int L,int R,int val,Tree tree[]){ if(L<=tree[root].l&&R>=tree[root].r) { tree[root].set=val; tree[root].add=0; tree[root].max=val; tree[root].min=val; tree[root].sum=(tree[root].r-tree[root].l+1)*val; return; } push_down(root,tree); int mid=(tree[root].l+tree[root].r)/2; if(L<=mid) update_set(root<<1,L,R,val,tree); if (R>mid) update_set(root<<1|1,L,R,val,tree); push_up(root,tree);}void query(int root,int L,int R,Tree tree[]){ if(L<=tree[root].l&&R>=tree[root].r) { _min=min(_min,tree[root].min); _max=max(_max,tree[root].max); _sum+=tree[root].sum; return; } push_down(root,tree); int mid=(tree[root].l+tree[root].r)/2; if(L<=mid) query(root<<1,L,R,tree); if(R>mid) query(root<<1|1,L,R,tree);}int main(){ int r,c,m; while (~scanf("%d%d%d",&r,&c,&m)) { int tp,x1,x2,y1,y2; for (int i=1;i<=r;i++) { build(1,1,c,big_tree[i]); } while (m--) { int v; scanf("%d%d%d%d%d",&tp,&x1,&y1,&x2,&y2); if (tp==1) { scanf("%d",&v); for (int i=x1;i<=x2;i++) { update_add(1,y1,y2,v,big_tree[i]); } } if (tp==2) { scanf("%d",&v); for (int i=x1;i<=x2;i++) { update_set(1,y1,y2,v,big_tree[i]); } } if (tp==3) { _max=0; _min=OO; _sum=0; for (int i=x1;i<=x2;i++) { query(1,y1,y2,big_tree[i]); } printf("%d %d %d\n",_sum,_min,_max); } } } return 0;}
- UVa 11992 - Fast Matrix Operations 成段更新,求最值与和
- Uva-11992-Fast Matrix Operations
- UVA 11992 Fast Matrix Operations
- Fast Matrix Operations ,uva 11992
- UVA 11992 Fast Matrix Operations
- UVA - 11992 Fast Matrix Operations
- UVa:11992 Fast Matrix Operations
- UVA 11992Fast Matrix Operations
- UVA 11992 - Fast Matrix Operations
- UVa 11992 Fast Matrix Operations
- uva 11992 Fast Matrix Operations
- UVA 11992 Fast Matrix Operations
- Uva 11992 Fast Matrix Operations
- UVA 11992 Fast Matrix Operations
- UVA 11992 Fast Matrix Operations (线段树区间更新)
- UVa 11992 Fast Matrix Operations / 线段树成段更新
- uva 11992 Fast Matrix Operations (线段树区间更新)
- uva 11992 Fast Matrix Operations(线段树)
- WINDOWS 同步
- js弹出窗口详解
- java中重定向和转发的差别
- 例题3-6
- #遍历E盘下的mp3文件
- UVa 11992 - Fast Matrix Operations 成段更新,求最值与和
- Liferay安装笔记(默认安装使用Mysql数据库)
- webkit 中 javascript 与 WebCore DOM 的绑定
- mysql master slave 搭建
- 为android程序添加背景音乐和Menu菜单
- 获取上一个页面的跳转地址
- C#开源资源大汇总
- 关于ORACLE的UPDATE更新多表的问题
- 转化为ico文件