zoukankan      html  css  js  c++  java
  • POJ——T 3067 Japan

    http://poj.org/problem?id=3067

    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 29474   Accepted: 7950

    Description

    Japan plans to welcome the ACM ICPC World Finals and a lot of roads must be built for the venue. Japan is tall island with N cities on the East coast and M cities on the West coast (M <= 1000, N <= 1000). K superhighways will be build. Cities on each coast are numbered 1, 2, ... from North to South. Each superhighway is straight line and connects city on the East coast with city of the West coast. The funding for the construction is guaranteed by ACM. A major portion of the sum is determined by the number of crossings between superhighways. At most two superhighways cross at one location. Write a program that calculates the number of the crossings between superhighways.

    Input

    The input file starts with T - the number of test cases. Each test case starts with three numbers – N, M, K. Each of the next K lines contains two numbers – the numbers of cities connected by the superhighway. The first one is the number of the city on the East coast and second one is the number of the city of the West coast.

    Output

    For each test case write one line on the standard output: 
    Test case (case number): (number of crossings)

    Sample Input

    1
    3 4 4
    1 4
    2 3
    3 2
    3 1

    Sample Output

    Test case 1: 5

    Source

     
     
    给坐标排序,求逆序对,手模一下好理解、
     1 #include <algorithm>
     2 #include <cstring>
     3 #include <cstdio>
     4 
     5 using namespace std;
     6 
     7 const int N(1111);
     8 int n,m,k,tr[N];
     9 struct Node
    10 {
    11     int west,east;
    12 }node[N*N];
    13 bool cmp(Node a,Node b)
    14 {
    15     if(a.east==b.east) return a.west>b.west;
    16     return a.east>b.east;
    17 }
    18 
    19 #define lowbit(x) (x&((~x)+1))
    20 inline void Update(int i,int x)
    21 {
    22     for(;i<=N;i+=lowbit(i)) tr[i]+=x;
    23 }
    24 inline int Query(int x)
    25 {
    26     int ret=0;
    27     for(;x;x-=lowbit(x)) ret+=tr[x];
    28     return ret;
    29 }
    30 
    31 int main()
    32 {
    33     int t; scanf("%d",&t); int h=1;
    34     for(long long ans=0;h<=t;h++,ans=0)
    35     {
    36         scanf("%d%d%d",&n,&m,&k);
    37         for(int i=1;i<=k;i++)
    38             scanf("%d%d",&node[i].west,&node[i].east);
    39         sort(node+1,node+k+1,cmp);
    40         memset(tr,0,sizeof(tr));
    41         for(int i=1;i<=k;Update(node[i++].west,1))
    42             ans+=(long long)Query(node[i].west-1);
    43         printf("Test case %d: %I64d
    ",h,ans);
    44     }
    45     return 0;
    46 }
    ——每当你想要放弃的时候,就想想是为了什么才一路坚持到现在。
  • 相关阅读:
    java.lang.ClassNotFoundException:org.springframework.web.context.ContextLoaderListener问题解决
    Unix
    Win7 扩容磁盘分区
    在Java中怎样高效的推断数组中是否包括某个元素
    2.JAVA编程思想——一切都是对象
    Cookie-Parser是怎样解析签名后的cookie的(同一时候对cookie和cookie-signature进行说明)
    hive 索引
    Javascript属性constructor/prototype的底层原理
    HTML5 的四个亮点
    零基础学python-1.5 第一个程序
  • 原文地址:https://www.cnblogs.com/Shy-key/p/7398080.html
Copyright © 2011-2022 走看看