hdu 4941 map结构体的用法

Magical Forest

Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 74    Accepted Submission(s): 31

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.

 

Sample Input

13 3 21 1 12 2 253 1 11 1 22 1 23 1 13 2 2

 

Sample Output

Case #1:121
Hint
No two fruits at the same location.
 

 

Author

UESTC

 

#include<stdio.h>#include<map>using namespace std;struct node{  int x,y;  bool friend operator<(node a,node b)  {    if(a.x!=b.x)       return a.x<b.x;    else       return a.y<b.y;  }}e;map<int ,int >m1,m2;map<node ,int >ma;int main(){    int t,n,m,k,n1,n2,x,y,c,q,op,tmp,i,cnt=0;    scanf("%d",&t);    while(t--)    {      scanf("%d%d%d",&n,&m,&k);      n1=n2=0;      for(i=1;i<=k;i++)      {         scanf("%d%d%d",&x,&y,&c);         if(m1[x]==0)            m1[x]=++n1;         if(m2[y]==0)             m2[y]=++n2;//给结点打上标签         e.x=m1[x];          e.y=m2[y];         ma[e]=c;//当前标签的值      }      scanf("%d",&q);      printf("Case #%d:\n",++cnt);      while(q--)      {       scanf("%d%d%d",&op,&x,&y);       if(op==1)       {         tmp=m1[x];//换当前两个值的标签         m1[x]=m1[y];         m1[y]=tmp;       }       else if(op==2)       {         tmp=m2[x];//同上         m2[x]=m2[y];         m2[y]=tmp;       }       else       {         e.x=m1[x];//根据标签求值         e.y=m2[y];         printf("%d\n",ma[e]);       }      }    }  return 0;}

Source

2014 Multi-University Training Contest 7