两边点连直线求交点总数 树状数组或线段树 poj 3067 Japan

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

Japan

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23602 Accepted: 6369

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

13 4 41 42 33 23 1

Sample Output

Test case 1: 5

Source

Southeastern Europe 2006

思路：树状数组
分析：
1 题目要求的是找到所有直线的交点总数，并且题目明确指出两条直线之间最多只有一个交点At most two superhighways cross at one location
2 我们应该先对这些直线进行排序：按照左边的编号从小到大，左边编号相同时按照右边编号从小到大。那么假设现在有一条直线1-3，那么能够和这条直线有交点的肯定是右边的编号大于3的，那么这个过程就可以利树状数组求和得到，求和完毕还要更新树状数组。
3 由于题目的k没有给定，那么最坏的情况1+2+...k

4 数据会超过int , 所以我们应该选择long long

代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int N = 1010;const int MAXN = 1000010;struct Node{    int x;    int y;    bool operator<(const Node& s)const{         if(x < s.x)              return true;         else if(x == s.x && y < s.y)              return true;         return false;    }};Node node[MAXN];int n , m , k;long long treeNum[N];int lowbit(int x){    return x&(-x);}long long getSum(int x){    long long sum = 0;    while(x){        sum += treeNum[x];        x -= lowbit(x);    }    return sum;}void add(int x , int val){    while(x < N){        treeNum[x] += val;        x += lowbit(x);    }}long long solve(){    long long ans = 0;    memset(treeNum , 0 , sizeof(treeNum));    sort(node , node+k);    for(int i = 0 ; i < k ; i++){        ans += i-getSum(node[i].y);        add(node[i].y , 1);    }    return ans;}int main(){    int cas = 1;    int Case;    scanf("%d" , &Case);    while(Case--){          scanf("%d%d%d" , &n , &m , &k);         for(int i = 0 ; i < k ; i++)             scanf("%d%d" , &node[i].x , &node[i].y);          printf("Test case %d: %lld\n" , cas++ , solve());    }    return 0;}

思路：线段树
分析：
1 题目要求的是找到所有直线的交点总数，并且题目明确指出两条直线之间最多只有一个交点
2 很明显我们应该先对这些直线进行排序：按照左边的编号从小到大，左边编号相同时按照右边编号从小到大。那么假设现在有一条直线1-3，那么能够和这条直线有交点的肯定是右边的编号大于3的，那么这个过程就可以利用线段树的查找，查找完毕还要更新线段树。
3 由于题目的n最大5*10^5，那么最坏的情况1+2+...n，会超过int , 所以我们应该选择long long.

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define MAXN 1000010int n , m , k;struct Point{   int x;   int y;};Point p[MAXN];struct Node{   int left;   int right;   int sum;};Node node[4*MAXN];bool cmp(Point a , Point b){   if(a.x < b.x)     return true;   else if(a.x == b.x && a.y < b.y)     return true;   return false;}void buildTree(int left , int right , int pos){   node[pos].left = left;   node[pos].right = right;   node[pos].sum = 0;   if(left == right)     return;   int mid = (left+right)>>1;   buildTree(left , mid , pos<<1);   buildTree(mid+1 , right , (pos<<1)+1);}int query(int left , int right , int pos){   if(node[pos].left == left && node[pos].right == right)     return node[pos].sum;   int mid = (node[pos].left + node[pos].right)>>1;   if(right <= mid)     return query(left , right , pos<<1);   else if(left > mid)     return query(left , right , (pos<<1)+1);   else     return query(left , mid , pos<<1)+query(mid+1 , right , (pos<<1)+1);}void update(int index , int pos){   if(node[pos].left == node[pos].right){      node[pos].sum++;      return;   }   int mid = (node[pos].left+node[pos].right)>>1;   if(index <= mid)     update(index , pos<<1);   else     update(index , (pos<<1)+1);   node[pos].sum = node[pos<<1].sum + node[(pos<<1)+1].sum;}int main(){  int t , Case = 1;  long long ans;//选择long long  scanf("%d" , &t);  while(t--){    scanf("%d%d%d" , &n , &m , &k);    for(int i = 0 ; i < k ; i++)       scanf("%d%d" , &p[i].x , &p[i].y);    sort(p , p+k , cmp);    buildTree(1 , m , 1);    ans = 0;    for(int i = 0 ; i < k ; i++){       if(p[i].y < m)         ans += query(p[i].y+1 , m , 1);       update(p[i].y , 1);    }    printf("Test case %d: %lld\n" , Case++ , ans);  }  return 0;}