ACM-ICPC 2018 徐州赛区网络预赛 - G Trace - 思维

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≤$(0<x0<x0<x,y\le 10000000y\le$ 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

3
1 4
4 1
3 3

样例输出

10

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

分析：

这是道让人想不到的思维题，实在解释不了其道理看博客吧传送门你们都什么脑子啊怎么想到的。。。太可怕了。。

code：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 50005;
int n;
ll X[N],Y[N];
ll solve(ll a[]){
    set<int> used;
    set<int>::iterator it;
    ll ans = 0;
    for(int i = n; i >= 1; i--){
        it = used.lower_bound(a[i]);
        if(it == used.begin()) ans += a[i];
        else{
            it--;
            ans += a[i] - *it;
        }
        used.insert(a[i]);
    }
    return ans;
}
int main(){
    scanf("%d",&n);
    for(int i = 1; i <= n; i++){
        scanf("%lld%lld",&X[i],&Y[i]);
    }
    ll ans = solve(X) + solve(Y);
    printf("%lld\n",ans);
    return 0;
}