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(树状数组) poj 3067
Japan
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 21769 | Accepted: 5885 |
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)
Test case (case number): (number of crossings)
Sample Input
13 4 41 42 33 23 1
Sample Output
Test case 1: 5
排个序，转化成逆序数
#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cstdlib>#include<cmath>#include<algorithm>using namespace std;#define N 2010int tt,n,m,k,c[N];struct node{ int x,y,pos;}e[N*N];bool cmp(node a,node b){ if(a.x!=b.x) return a.x<b.x; return a.y<b.y;};int lowbit(int x){ return x&(-x);}void update(int pos,int m){ while(pos<N) { c[pos]+=m; pos+=lowbit(pos); }}int sum(int x){ int sum=0; while(x>0) { sum+=c[x]; x-=lowbit(x); } return sum;}int main(){ int cas=1; scanf("%d",&tt); while(tt--) { long long ans=0; memset(c,0,sizeof(c)); scanf("%d%d%d",&n,&m,&k); for(int i=1;i<=k;i++) scanf("%d%d",&e[i].x,&e[i].y); sort(e+1,e+1+k,cmp); for(int i=1;i<=k;i++) { update(e[i].y,1); ans+=i-sum(e[i].y); } printf("Test case %d: %I64d\n",cas,ans); cas++; } return 0;}
(树状数组) poj 3067
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