Japan-(数据结构树状数组)

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)

        题目大意就是：东部西部有两个岛，岛上分别有N个城市和M个城市，现在修建高速公路，求交点数。

        解题方法：树状数组、逆序列。

        谈谈这道题的体会：

        在刚刚弄懂Lowbit(x)的神奇之处，迫不及待的准备小试身手，于是遇到了这道题。

        （传送门Lowbit(t）https://www.cnblogs.com/hsd-/p/6139376.html 

                                       http://www.cnblogs.com/dilthey/p/6804136.html  )

        刚开始的时候思路一点也不清晰，甚至是不理解这道题和树状数组有什么关系。只是稿纸上画图找寻规律。

        根据题意我们可以将题目化简，考虑一下题目的转折点->就是什么时候会出现交点。经过思考可以发现，在N是升序排列的前提之下，我现在讨论的(假设是第k组数据） 只要存在a[k].M 小于 a[1].M~a[k-1].M中的数就会有交点，且有几个小于就有几个交点。由此我们便可以想到用树状数组(既每次放入的点有可能会影响到之后输出的结果。因为对树状数组的领悟很浅所以想了半天他们之间联系，写了一大张演草纸==)来储存a[MAX_K].M的信息，最后再用逆序列的思想就可解决啦。

上代码

#include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define lowbit(x) (x)&(-x) typedef long long ll; int T,N,M,K; ll c[1005]; struct node     {         int N,M;     }a[1005*1005]; bool cmp(node a,node b ) {     if(a.N==b.N) return a.M<b.M;     return a.N<b.N; } ll sum(int a) {     ll ans=0;     for(int i=a;i>0;i-=lowbit(i)){         ans+=c[i];     }     return ans; } void add(int a) {     while(a<=M){         c[a]=c[a]+1;         a+=lowbit(a);     } } int main() {     scanf("%d",&T);     int t=1;     while(T--)     {         scanf("%d%d%d",&N,&M,&K);         for(int i=1;i<=K;i++){             scanf("%d%d",&a[i].N,&a[i].M);         }         sort(a+1,a+1+K,cmp);         memset(c,0,sizeof(c));         ll res=0;         for(int i=1;i<=K;i++)         {             add(a[i].M);            // cout<<sum(a[i].M)<<endl;             res+=(i-sum(a[i].M));//逆序列思想             //cout<<sum<<endl;         }         printf("Test case %d: %I64d\n",t++,res);     } }

当！就是这样！