POJ 3067 Japan

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POJ 3067 Japan

唔~~~咋一看之下还以为要开二维数组，不过倒是看得出这题要么用树状数组要么用线段树（居然可以用表情！！！）咳咳。。。区间问题嘛~后来想了想，似乎不是这么回事，在脑子里想象一下，发现这题可以转换成一维树状数组求逆序数~

先对n从小到大排序，如果相等，则对m进行从小到大排序，唔，对道路们排序不影响结果这是必然的，然后对m求逆序数(画出来就知道，其实这样就能很清晰直观的用焦点个数来判断了~(j>i)mj<mi的话这两条道路必有交点，以此类推)~~就TLE了！！！！~~~~哎~~不知道有木有大神有更快的方法~~懒得想，用了快速io~~wa~~~~~~~~才发现最后的结果是long型的好烦，总是不会去考虑数据类型，等着wa了才会去找~~~眼拙~~在题目里没找着k的范围~~~！

AC代码：

import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;import java.io.OutputStreamWriter;import java.io.PrintWriter;import java.io.StreamTokenizer;import java.util.Arrays;import java.util.Comparator;class way{int x,y;}public class Main{static StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));public static int nextInt()throws IOException {in.nextToken();return (int)in.nval;}private static int lowbit(int x) {return x&(-x);}private static void upDate(int[] c,int n,int x,int add){for(int i=x;i<=n;i+=lowbit(i))c[i]+=add;}private static int sum(int[] c,int x){int sum=0;for(int i=x;i>0;i-=lowbit(i))sum+=c[i];return sum;}public static void main(String[] args)throws IOException{int t=nextInt();for(int i=1;i<=t;i++){int n=nextInt();int m=nextInt();int k=nextInt();way xy[]=new way[k+1];for(int j=1;j<=k;j++){xy[j]=new way();xy[j].x=nextInt();xy[j].y=nextInt();}/*if(n==0 && m==0){out.println("Test case "+i+": "+0);continue;}*/Arrays.sort(xy,1,k+1,new Comparator<way>() {@Overridepublic int compare(way a, way b){if(a.x!=b.x)return (a.x>b.x)?1:-1;return (a.y>b.y)?1:-1;}});int max=n>=m?n:m;int c[]=new int[max+1];long sum=0;for(int j=1;j<=k;j++){sum+=j-1-sum(c,xy[j].y);upDate(c,max,xy[j].y,1);}out.println("Test case "+i+": "+sum);}out.flush();}}

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