hdoj 4941 Magical Forest 【STL map】

Magical Forest Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1287    Accepted Submission(s): 580 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.

题意：一个n*m的地图，上面有k个物品，现在已经给出每个物品的位置以及对应的价值。下面有Q次操作 op x y

op为1时，表示把第x行和第y行交换；

op为2时，表示把第x列和第y列交换；

op为3时，查询(x, y)物品的价值，没有物品输出0

思路：由于n和m过大，不能直接存储。可以使用map建立映射，然后直接对map映射值进行交换就可以了。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
map<int, int> X, Y;
map<int, map<int, int> > val;
int kcase = 1;
void solve()
{
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    X.clear(), Y.clear(), val.clear();
    int x, y ,v, op;
    int xnum = 0, ynum = 0;
    for(int i = 0; i < k; i++)
    {
        scanf("%d%d%d", &x, &y, &v);
        if(X[x] == 0)
            X[x] = ++xnum;
        if(Y[y] == 0)
            Y[y] = ++ynum;
        val[X[x]][Y[y]] = v;
    }
    int Q;
    scanf("%d", &Q);
    printf("Case #%d:\n", kcase++);
    while(Q--)
    {
        scanf("%d%d%d", &op, &x, &y);
        if(op == 1)
            swap(X[x], X[y]);
        else if(op == 2)
            swap(Y[x], Y[y]);
        else
            printf("%d\n", val[X[x]][Y[y]]);
    }
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--){
        solve();
    }
    return 0;
}