zoukankan      html  css  js  c++  java
  • hdu 4941 2014 Multi-University Training Contest 7 1007

    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。。。
     1 #include<iostream>
     2 #include<cstring>
     3 #include<cstdlib>
     4 #include<cstdio>
     5 #include<algorithm>
     6 #include<cmath>
     7 #include<queue>
     8 #include<map>
     9 //#include<pair>
    10 
    11 #define N 1505
    12 #define M 105
    13 #define mod 1000000007
    14 #define mod2 100000000
    15 #define ll long long
    16 #define maxi(a,b) (a)>(b)? (a) : (b)
    17 #define mini(a,b) (a)<(b)? (a) : (b)
    18 
    19 using namespace std;
    20 
    21 map<pair<int,int>,int>v;
    22 map<int,int>X,Y;
    23 int n,m,k;
    24 int q,a,b;
    25 int x,y,c;
    26 int tt;
    27 int T;
    28 
    29 int main()
    30 {
    31   //  freopen("data.in","r",stdin);
    32     scanf("%d",&T);
    33     for(int cnt=1;cnt<=T;cnt++)
    34     //while(T--)
    35     //while(scanf("%d%d",&n,&m)!=EOF)
    36     {
    37      //
    38         printf("Case #%d:
    ",cnt);
    39         v.clear();X.clear();Y.clear();
    40         scanf("%d%d%d",&n,&m,&k);
    41         while(k--){
    42             scanf("%d%d%d",&x,&y,&c);
    43             v[make_pair(x,y)]=c;
    44            // printf(" %d %d %d
    ",x,y,c);
    45             X[x]=x;
    46             Y[y]=y;
    47            // printf(" %d:%d %d:%d
    ",x,X[x],y,Y[y]);
    48         }
    49         scanf("%d",&tt);
    50         while(tt--)
    51         {
    52             scanf("%d%d%d",&q,&a,&b);
    53             if(q==1){
    54                 swap(X[a],X[b]);
    55                 //printf(" %d:%d %d:%d
    ",a,X[a],b,X[b]);
    56             }
    57             else if(q==2){
    58                 swap(Y[a],Y[b]);
    59                // printf(" %d:%d %d:%d
    ",a,Y[a],b,Y[b]);
    60             }
    61             else{
    62                 printf("%d
    ",v[make_pair(X[a],Y[b])]);
    63             }
    64 
    65         }
    66 
    67 
    68 
    69     }
    70 
    71 
    72     return 0;
    73 }
  • 相关阅读:
    Qt 智能指针学习(7种指针)
    Springmvc+Spring+Hibernate搭建方法
    DDD分层架构之领域实体(基础篇)
    LeetCode
    RabbitMQ
    Cocos2d-x环境搭建
    使用快捷键提升C#开发效率
    C# 6.0
    avalonjs 1.3.7发布
    VS2015安装
  • 原文地址:https://www.cnblogs.com/njczy2010/p/3908262.html
Copyright © 2011-2022 走看看