Magical Forest

Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 134    Accepted Submission(s): 69

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
 
Recommend
We have carefully selected several similar problems for you:  4944 4943 4942 4941 4940 
 
 
思路:stl。。。
 #include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
//#include<pair> #define N 1505
#define M 105
#define mod 1000000007
#define mod2 100000000
#define ll long long
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b) using namespace std; map<pair<int,int>,int>v;
map<int,int>X,Y;
int n,m,k;
int q,a,b;
int x,y,c;
int tt;
int T; int main()
{
// freopen("data.in","r",stdin);
scanf("%d",&T);
for(int cnt=;cnt<=T;cnt++)
//while(T--)
//while(scanf("%d%d",&n,&m)!=EOF)
{
//
printf("Case #%d:\n",cnt);
v.clear();X.clear();Y.clear();
scanf("%d%d%d",&n,&m,&k);
while(k--){
scanf("%d%d%d",&x,&y,&c);
v[make_pair(x,y)]=c;
// printf(" %d %d %d\n",x,y,c);
X[x]=x;
Y[y]=y;
// printf(" %d:%d %d:%d\n",x,X[x],y,Y[y]);
}
scanf("%d",&tt);
while(tt--)
{
scanf("%d%d%d",&q,&a,&b);
if(q==){
swap(X[a],X[b]);
//printf(" %d:%d %d:%d\n",a,X[a],b,X[b]);
}
else if(q==){
swap(Y[a],Y[b]);
// printf(" %d:%d %d:%d\n",a,Y[a],b,Y[b]);
}
else{
printf("%d\n",v[make_pair(X[a],Y[b])]);
} } } return ;
}

hdu 4941 2014 Multi-University Training Contest 7 1007的更多相关文章

  1. hdu 4915 Parenthese sequence--2014 Multi-University Training Contest 5

    主题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4915 Parenthese sequence Time Limit: 2000/1000 MS (Ja ...

  2. hdu 4902 Nice boat--2014 Multi-University Training Contest 4

    题目链接:http://acm.hdu.edu.cn/showproblem.php? pid=4902 Nice boat Time Limit: 30000/15000 MS (Java/Othe ...

  3. hdu 4925 Apple Tree--2014 Multi-University Training Contest 6

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4925 Apple Tree Time Limit: 2000/1000 MS (Java/Others ...

  4. HDU校赛 | 2019 Multi-University Training Contest 6

    2019 Multi-University Training Contest 6 http://acm.hdu.edu.cn/contests/contest_show.php?cid=853 100 ...

  5. HDU校赛 | 2019 Multi-University Training Contest 5

    2019 Multi-University Training Contest 5 http://acm.hdu.edu.cn/contests/contest_show.php?cid=852 100 ...

  6. HDU校赛 | 2019 Multi-University Training Contest 4

    2019 Multi-University Training Contest 4 http://acm.hdu.edu.cn/contests/contest_show.php?cid=851 100 ...

  7. HDU校赛 | 2019 Multi-University Training Contest 3

    2019 Multi-University Training Contest 3 http://acm.hdu.edu.cn/contests/contest_show.php?cid=850 100 ...

  8. HDU校赛 | 2019 Multi-University Training Contest 2

    2019 Multi-University Training Contest 2 http://acm.hdu.edu.cn/contests/contest_show.php?cid=849 100 ...

  9. HDU校赛 | 2019 Multi-University Training Contest 1

    2019 Multi-University Training Contest 1 http://acm.hdu.edu.cn/contests/contest_show.php?cid=848 100 ...

随机推荐

  1. 01_3Java Application初步

    01_3Java Application初步 l Java源文件以“java”为扩展名.源文件的基本组成部分是类(class),如本例中的HelloWorld类. l 一个源文件中最多只有一个publ ...

  2. iOS 开发 Xib 的嵌套使用

    最近公司项目需要使用 Xib 中嵌套 Xib来布局界面的, 研究了很久才实现!!! 分享给大家,希望帮助到更多的开发者...... 开发中自定义界面有两种方式 一: 纯代码实现 适合单个极度复杂的界面 ...

  3. @private@protected@public@package

    @private@protected@public@package 为了强制一个对象隐藏其数据,编译器限制实例变量范围以限制其在程序中的可见性 但是为了提供灵活性,苹果也让开发者显式设置范围(四选一) ...

  4. 【转】LDA-linear discriminant analysis

    分类问题也可以用降维来理解,比如一个D维的数据点x,我们可以采用下面的映射进行线性的降维, y=θTx 在计算出y后,就可以选择一个阈值h,来进行分类.正如我们在前面的PCA模型中看到的,降维会有信息 ...

  5. 笔记--Day2--python基础2

    一.鸡汤 1.提高自我修养 2.人丑就要多读书 3.多走走,开拓眼界 二.目录: 1.列表.元组操作 2.字符串操作 3.字典操作 dict是无序的 key必须是唯一的 4.集合操作 集合是一个无序的 ...

  6. Windows10 关闭自动更新

    win+R调出运行窗口: 输入services.msc,查找 跳出服务窗口,点击windows update设置禁用即可 Windows Update Medic Service没办法禁用,需要采用其 ...

  7. 【windows】win7 sp1 系统语言中英文切换

    注:Windows 7 Ultimate and Windows 7 Enterprise (旗舰版和企业版) 可以直接在控制面板/地区和语言中修改显示语言,其他系统不行 进入网站下载相关的MUI包安 ...

  8. GoF23种设计模式之结构型模式之代理模式

    一.概述 为其他对象提供一种代理以控制对这个对象的访问. 二.适用性 1.远程代理(RemoteProxy):为一个对象在不同的地址空间土工局部代表. 2.虚代理(VirtualProxy):根据需要 ...

  9. Python学习笔记(一):基础知识

    一.什么是python? python是一种面向对象.解释型的计算机语言,它的特点是语法简洁.优雅.简单易学 二.编译型语言和解释型语言 编译型语言就是把程序编译成计算机语言然后执行,(一次编译到处运 ...

  10. CodeForces 8D Two Friends 判断三个圆相交

    题意: 有两个人\(Alan\)和\(Bob\),他们现在都在\(A\)点,现在\(Bob\)想去\(B\)点,\(Alan\)想先到\(C\)点再去\(B\)点. \(Alan\)所走的总路程不能超 ...