map应用——HDU 4941

最新推荐文章于 2020-01-07 16:10:45 发布

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本文介绍了一种高效处理二维网格中元素查询与位置交换的方法。针对动态变化的N*M网格，实现快速响应行列交换操作及特定位置的能量值查询。利用map和pair数据结构优化存储与检索效率。

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对应HDU题目：点击打开链接

Magical Forest

Time Limit: 24000/12000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 947 Accepted Submission(s): 427

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.

map+pair，看了别人的才知道可以这么用，太年轻了。。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<map>
using namespace std;
#define N 100010
#define mp(x,y) make_pair(x,y)
int T,n;
map<int,int> r,c;
map<pair<int,int>,int> ans;

int main()
{
    //freopen("in.txt","r",stdin);
    int cas=1,i,x,y,z;
    scanf("%d",&T);
    while(T--)
    {
        printf("Case #%d:\n",cas++);
        scanf("%d%d%d",&n,&n,&n);
        r.clear();
        c.clear();
        ans.clear();
        for(i=1; i<=n; i++){
            scanf("%d%d%d", &x,&y,&z);
            r[x]=x;
            c[y]=y;
            ans[mp(x,y)]=z;
        }
		int tmp;
        scanf("%d", &n);
        while(n--)
        {
            scanf("%d%d%d", &z,&x,&y);
			switch(z)
			{
				case 1:{
					/*r[x]=r[x]^r[y];
					r[y]=r[x]^r[y];
					r[x]=r[x]^r[y];*/
					tmp=r[x]; r[x]=r[y]; r[y]=tmp;//这里如果换成上面注释的就不行，^运算两个数不可以相等，这里r[x]可能==r[y]
					break;}
				case 2:{
					/*c[x]=c[x]^c[y];
					c[y]=c[x]^c[y];
					c[x]=c[x]^c[y];*/
					tmp=c[x]; c[x]=c[y]; c[y]=tmp;
					break;}
				case 3:{
					pair<int,int> b=mp(r[x],c[y]);
                	printf("%d\n",ans[b]);}
			}
        }
    }
    return 0;
}