HDOJ 4941 Magical Forest

2024-05-05 12:38

文章标签 forest hdoj magical 4941

本文主要是介绍HDOJ 4941 Magical Forest，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目:

Magical Forest

Time Limit: 24000/12000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 389 Accepted Submission(s): 186

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

传送门：点击打开链接

解题思路：

离散化。我们可以存储所有给出的点，对所有的点的行和列进行离散化。可以用map来进行映射，定义mpr表示mpr[i]表示现在第i行是原来的mpr[i]行。在进行交换操作时，只需要将mpr和mpc的值进行交换，查找时进行二分查找。

代码：

#include <cstdio>
#include <map>
#include <cstring>
#include <algorithm>
using namespace std;const int MAXN = 1e5+10;
struct Forest
{int r, c, v;
};
Forest p[MAXN];
map<int, int> mpr;
map<int, int> mpc;bool cmp(const Forest &a, const Forest &b)
{return a.r==b.r ? a.c<b.c : a.r<b.r;
}int find(int row, int col, int k)
{row = 0==mpr[row] ? row : mpr[row];col = 0==mpc[col] ? col : mpc[col];int l = 0, r = k-1;while(l <= r){int mid = (l+r) >> 1;if(p[mid].r==row && p[mid].c==col)return p[mid].v;if(p[mid].r<row || (p[mid].r==row && p[mid].c<col))l = mid + 1;elser = mid - 1;}return 0;
}int main()
{int icase, n, m, t, k;while(~scanf("%d", &icase)){int w = 1;while(icase--){mpr.clear(); mpc.clear();printf("Case #%d:\n", w++);scanf("%d%d%d", &n, &m, &k);for(int i = 0; i < k; ++i){scanf("%d%d%d", &p[i].r, &p[i].c, &p[i].v);//  mpr[p[i].r] = p[i].r; mpc[p[i].c] = p[i].c;}sort(p, p+k, cmp);scanf("%d", &t);while(t--){int q, a, b;scanf("%d%d%d", &q, &a, &b);switch(q){case 1:{int t1 = 0==mpr[a] ? a : mpr[a];int t2 = 0==mpr[b] ? b : mpr[b];mpr[a] = t2;mpr[b] = t1;break;}case 2:{int t1 = 0==mpc[a] ? a : mpc[a];int t2 = 0==mpc[b] ? b : mpc[b];mpc[a] = t2;mpc[b] = t1;break;}case 3:{printf("%d\n", find(a, b, k));break;}}}}}return 0;
}

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