STL ： map函数的运用 --- hdu 4941 : Magical Forest

Magical Forest

Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 724    Accepted Submission(s): 343

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

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

 

Sample Output

Case #1: 1 2 1

Hint

No two fruits at the same location.

 

Author

UESTC

 

Source

2014 Multi-University Training Contest 7

 

---

Mean:

有一片N*M的森林，里面种着一些果实，现在有三种操作：1.交换第i行和第j行的果实；2.交换第i列和第j列的果实；3.查询第i行第j列有多少果实。

 

analyse:

由于地图的范围达到了2*1e9，我们不可能用二维数组，必须离散化。

我们用三个map函数，一个存行坐标，一个存列坐标，另一个存果实的数目。在变换的时候，只需交换行坐标的map或者列坐标的map,存果实的map函数是一直不变的，这样就大大减少了时间的消耗。

 

Time complexity：O(k)

Source code：

//Memory   Time
// 1347K   0MS
// by : Snarl_jsb
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define MAX 1100
#define LL long long
using namespace std;
map<int,int>row;
map<int,int>col;
map<int,map<int,int> >val;
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("cin.txt","r",stdin);
    #endif
    int T,kase=1;
    cin>>T;
    while(T--)
    {
        row.clear();
        col.clear();
        val.clear();
        int x,y,k,c;
        int rrow=1,ccol=1;
        scanf("%d %d %d",&x,&y,&k);
        while(k--)
        {
            scanf("%d %d %d",&x,&y,&c);
            if(!row[x])
                row[x]=rrow++;
            if(!col[y])
                col[y]=ccol++;
            x=row[x],y=col[y];
            val[x][y]=c;
        }
        printf("Case #%d:
",kase++);
        int Q,tmp;
        cin>>Q;
        while(Q--)
        {
            scanf("%d %d %d",&c,&x,&y);
            if(c==1)
            {
                tmp=row[x];
                row[x]=row[y];
                row[y]=tmp;
            }
            else if(c==2)
            {
                tmp=col[x];
                col[x]=col[y];
                col[y]=tmp;
            }
            else
                printf("%d
",val[row[x]][col[y]]);
        }
    }
    return 0;
}