Problem Description
There is a forest can be seen as N * M grid. In this forest, there is some magical fruits, These fruits can provide a lot of energy, Each fruit has its location(Xi, Yi) and the energy can be provided Ci.
However, the forest will make the following change sometimes:
1. Two
rows of forest exchange.
2. Two columns of forest exchange.
Fortunately,
two rows(columns) can exchange only if both of them contain fruits or none of
them contain fruits.
Your superior attach importance to these magical
fruit, he needs to know this forest information at any time, and you as his best
programmer, you need to write a program in order to ask his answers quick every
time.
Input
The input consists of multiple test cases.
The
first line has one integer W. Indicates the case
number.(1<=W<=5)
For each case, the first line has three integers
N, M, K. Indicates that the forest can be seen as maps N rows, M columns, there
are K fruits on the map.(1<=N, M<=2*10^9, 0<=K<=10^5)
The
next K lines, each line has three integers X, Y, C, indicates that there is a
fruit with C energy in X row, Y column. (0<=X<=N-1, 0<=Y<=M-1,
1<=C<=1000)
The next line has one integer T.
(0<=T<=10^5)
The next T lines, each line has three integers Q, A, B.
If Q = 1 indicates that this is a map of the row switching operation, the A
row and B row exchange.
If Q = 2 indicates that this is a map of the column
switching operation, the A column and B column exchange.
If Q = 3 means that
it is time to ask your boss for the map, asked about the situation in (A, B).
(Ensure that all given A, B are legal. )
Output
For each case, you should output "Case #C:" first, where C indicates the case number and counts from 1.
In each case, for
every time the boss asked, output an integer X, if asked point have fruit, then
the output is the energy of the fruit, otherwise the output is 0
input
1
3 3 2
1 1 1
2 2 2
5
3 1 1
1 1 2
2 1 2
3 1 1
3 2 2
output
Case #1:
1
2
1
考察离散化的模拟题。
1 #include <iostream>
2 #include <cstdio>
3 #include <map>
4 #include <algorithm>
5
6 using namespace std;
7
8 struct Node
9 {
10 int x;
11 int y;
12 int c;
13 }l[100000];
14
15 map <int, map<int, int> > fmap;
16 map <int ,int>hashx,hashy;
17 int T,n,m,k,mapx,mapy,t,cas,lianx[100010],liany[100010],Q,A,B,X,Y;
18
19 bool cmpx(Node a,Node b)
20 {
21 return a.x<b.x;
22 }
23 bool cmpy(Node a,Node b)
24 {
25 return a.y<b.y;
26 }
27
28 int main()
29 {
30 int CAS;
31 scanf("%d", &CAS);
32 for(int cas=1;cas<=CAS;cas++)
33 {
34 scanf("%d%d%d",&n,&m,&k);
35 printf("Case #%d:\n",cas);
36 for(int i=0;i<k;i++)scanf("%d%d%d",&l[i].x,&l[i].y,&l[i].c);
37 fmap.clear();hashx.clear();hashy.clear();mapx=mapy=0;
38 sort(l,l+k,cmpx);
39 for(int i=0;i<k;i++)if(!hashx[l[i].x])hashx[l[i].x]=++mapx;
40 sort(l,l+k,cmpy);
41 for(int i=0;i<k;i++)
42 {
43 if(!hashy[l[i].y])hashy[l[i].y]=++mapy;
44 fmap[hashx[l[i].x]][hashy[l[i].y]]=l[i].c;
45 }
46 for(int i=1;i<=mapx;i++)lianx[i]=i;
47 for(int i=1;i<=mapy;i++)liany[i]=i;
48 scanf("%d",&t);
49 for(int i=0;i<t;i++)
50 {
51 scanf("%d%d%d",&Q,&A,&B);
52 if(Q==1)
53 {
54 X=hashx[A];Y=hashx[B];
55 if(X && Y)swap(lianx[X],lianx[Y]);
56 }
57 else if(Q==2)
58 {
59 X=hashy[A];Y=hashy[B];
60 if(X && Y)swap(liany[X],liany[Y]);
61 }
62 else if(Q==3)
63 {
64 X=hashx[A];Y=hashy[B];
65 if(X && Y)printf("%d\n",fmap[lianx[X]][liany[Y]]);
66 else printf("0\n");
67 }
68 }
69 }
70 return 0;
71 }