Trace 2018徐州icpc网络赛（二分）（树状数组）

Trace

There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xxx , yyy ) means the wave is a rectangle whose vertexes are ( 000 , 000 ), ( xxx , 000 ), ( 000 , yyy ), ( xxx , yyy ). Every time the wave will wash out the trace of former wave in its range and remain its own trace of ( xxx , 000 ) -> ( xxx , yyy ) and ( 000 , yyy ) -> ( xxx , yyy ). Now the toad on the coast wants to know the total length of trace on the coast after n waves. It's guaranteed that a wave will not cover the other completely.

Input

The first line is the number of waves n(n≤50000)n(n \le 50000)n(n≤50000).

The next nnn lines，each contains two numbers xxx yyy ,( 0<x0 < x0<x , y≤10000000y \le 10000000y≤10000000 )，the iii-th line means the iii-th second there comes a wave of ( xxx , yyy ), it's guaranteed that when 1≤i1 \le i1≤i , j≤nj \le nj≤n ，xi≤xjx_i \le x_jxi≤xj and yi≤yjy_i \le y_jyi≤yj don't set up at the same time.

Output

An Integer stands for the answer.

Hint：

As for the sample input, the answer is 3+3+1+1+1+1=103+3+1+1+1+1=103+3+1+1+1+1=10

样例输入复制

样例输出复制

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

比赛的时候想的是手撸一个平衡树排序，然后二分查找这个点左右两边的点，再比较他们的边值，结果数据结构写炸了

赛后看到网上题解才想起set和list 这些stl 太弱了Orz

用set这类的话就把一个点的边值拆成两个处理，从后往前，（后面的点总是覆盖前面），看看当今的点时候比后面的点边值高出多少

set的二分还是要%一下，毕竟没这么写过

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 int n;

 const int maxn = 5e4+;

 const int maxnn = 1e7+;

 vector<int>vec1,vec2;

 ll solve(vector<int>vec)

 {

     ll ans = ;

     int sz = vec.size();

     set<int>st;

     for(int i=n-;i>=;i--)

     {

         set<int>::iterator it = st.lower_bound(vec[i]);

         if(it == st.begin())ans += vec[i];

         else

         {

             it--;

             ans += vec[i] - *it;

         }

         st.insert(vec[i]);

     }

     return ans;

 }

 int main()

 {

     scanf("%d",&n);

     int x,y;

     for(int i=;i<=n;i++)

     {

      scanf("%d%d",&x,&y);

      vec1.push_back(x);

      vec2.push_back(y);

     }

     printf("%lld\n",solve(vec1)+solve(vec2));

 }

除了可以二分之外还有树状数组的写法，能树状数组肯定可以线段树。这里先补上树状数组的做法。

trex表示是一个记录相应x位置的y的值的前缀和，trey也类似

　　那么插入（4，1）的时候更改trex，trey， y from 0 到 1 所遇到的横向遇到x都是3；x from 0 to 3 和 from 4 to 4 对应y值不同，这样就用到树状数组的区间更新

　　把query 出来的上一个位置+1的位置更新加上现在的点和之前的差值然后在现在位置+1的位置减去例如：trex（x在各个位置对应的y值） 3+1位置加上差值（1-3） 4+1位置加上差值的相反数

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 const int maxn = 5e4+;

 const int maxx = 1e7+;

 int n;

 struct Node

 {

     int x,y;

 }node[maxn];

 int trex[maxx];

 int trey[maxx];

 int maxxx = ;

 int lowbit(int x)

 {

     return x &(-x);

 }

 void add(int x,int val,int *tre)

 {

     for(int i=x;i<=maxxx;i+=lowbit(i))

     {

         tre[i] += val;

     }

 }

 int query(int x,int *tre)

 {

     int ans = ;

     for(int i=x;i>;i-=lowbit(i))

     {

         ans += tre[i];

     }

     return ans;

 }

 int main()

 {

     scanf("%d",&n);

     for(int i=;i<=n;i++)

     {

         scanf("%d%d",&node[i].x,&node[i].y);

         maxxx = max(maxxx,max(node[i].x,node[i].y));

     }

     ll ans = ;

     for(int i=n;i>;i--)

     {

         int x = node[i].x;

         int y = node[i].y;

         int lenx = query(y,trey);

         int leny = query(x,trex);

         ans += x - lenx;

         ans += y - leny;

         int valy = query(x,trex);

         int valx = query(y,trey);

         add(valy+,x-valx,trey);

         add(y+,valx-x,trey);

         add(valx+,y-valy,trex);

         add(x+,valy-y,trex);

         //printf("%d---%d    %d---%d    %d----%d\n",x,y,x-valx,y-valy,valx,valy);

     }

     printf("%lld\n",ans);

 }