【hdu4941】hash解决大行列交换问题~

最新推荐文章于 2022-04-27 00:45:00 发布

Sky-J

最新推荐文章于 2022-04-27 00:45:00 发布

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分类专栏： ACM-胡搞

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本文详细介绍了如何通过编程解决魔法森林中的水果位置变动问题，实现快速查询并进行行列交换操作。通过案例分析，展示了输入输出规范，以及如何在多轮操作中维护森林地图的更新，最终实现高效的数据管理和查询。

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Magical Forest

Time Limit: 24000/12000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 350 Accepted Submission(s): 165

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

Sample Output

      
       Case #1:
1
2
1


       
        
         Hint
        No two fruits at the same location.

Author

UESTC

Source

2014 Multi-University Training Contest 7

几个map一起上就A了。。。。

先对行map再对列map，交换直接交换map的值，最后结果是原来的的那个行和列，对原来的那个行列也map下，查询就行了

#define DeBUG
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <string>
#include <set>
#include <sstream>
#include <map>
#include <list>
#include <bitset>
using namespace std ;
#define zero {0}
#define INF 0x3f3f3f3f
#define EPS 1e-6
#define TRUE true
#define FALSE false
typedef long long LL;
const double PI = acos(-1.0);
//#pragma comment(linker, "/STACK:102400000,102400000")
inline int sgn(double x)
{
    return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);
}
#define N 100005
std::map<int, int> mprow;
std::map<int, int> mpcol;
std::map<int, map<int, int> > mp;
struct Node
{
    int x, y;
    int c;
};
Node node[N];
int cnt=1;
int main()
{
#ifdef DeBUGs
    freopen("C:\\Users\\Sky\\Desktop\\1.in", "r", stdin);
#endif
    int T;
    scanf("%d", &T);
    while (T--)
    {
        printf("Case #%d:\n",cnt++);
        memset(node, 0, sizeof(node));
        mprow.clear();
        mpcol.clear();
        mp.clear();
        int n, m, k;
        int q;
        scanf("%d%d%d", &n, &m, &k);
        for (int i = 0; i < k; i++)
        {
            scanf("%d%d%d", &node[i].x, &node[i].y, &node[i].c);
            node[i].x++;
            node[i].y++;
            mprow[node[i].x] = node[i].x;
            mpcol[node[i].y] = node[i].y;
            mp[node[i].x][node[i].y] = node[i].c;
        }
        scanf("%d", &q);
        int op, x, y;
        int t;
        for (int i = 0; i < q; i++)
        {
            scanf("%d%d%d", &op, &x, &y);
            x++;
            y++;
            if (op == 1)
            {
                if (mprow[x] != 0 && mprow[y] != 0)
                {
                    t = mprow[x];
                    mprow[x] = mprow[y];
                    mprow[y] = t;
                }
            }
            else if (op == 2)
            {
                if (mpcol[x] != 0 && mpcol[y] != 0)
                {
                    t = mpcol[x];
                    mpcol[x] = mpcol[y];
                    mpcol[y] = t;
                }
            }
            else
            {
                int nowx,nowy;
                nowx=mprow[x];
                nowy=mpcol[y];
                printf("%d\n", mp[nowx][nowy]);
            }
        }
    }
    return 0;
}