## ACM-ICPC 2018 徐州赛区网络预赛 G. Trace (思维，贪心)

Trace

• 35.78%
• 1000ms
• 262144K

There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xx , yy ) means the wave is a rectangle whose vertexes are ( 00 , 00 ), ( xx , 00 ), ( 00 , yy ), ( xx , yy ). Every time the wave will wash out the trace of former wave in its range and remain its own trace of ( xx , 00 ) -> ( xx , yy ) and ( 00 , yy ) -> ( xx , yy ). 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 \le 50000)n(n≤50000).

The next nn lines，each contains two numbers xx yy ,( 0 < x0<x , y \le 10000000y≤10000000 )，the ii-th line means the ii-th second there comes a wave of ( xx , yy ), it's guaranteed that when 1 \le i1≤i , j \le njn ，x_i \le x_jxixj and y_i \le y_jyiyj 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=10

#### 样例输入复制

3
1 4
4 1
3 3

#### 样例输出复制

10

#### 题目来源

ACM-ICPC 2018 徐州赛区网络预赛

## 题意：

n波矩形海浪，每次都会在沙滩上留下痕迹，求最后的痕迹长度。

## 代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}

inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int x, y;
set<int> sx, sy;
pii a[maxn];
int main()
{
//freopen("D:\\code\\text\\input.txt","r",stdin);
//freopen("D:\\code\\text\\output.txt","w",stdout);
int n;
gbtb;
cin >> n;
ll ans = 0ll;
sx.insert(0);
sy.insert(0);

repd(i, 1, n)
{
cin >> x >> y;
a[i].fi = x;
a[i].se = y;
}
for (int i = n; i >= 1; --i)
{
x = a[i].fi;
y = a[i].se;

auto it = sx.upper_bound(x);
it--;
ans += 1ll * (x - (*it));

it = sy.upper_bound(y);
it--;
ans += 1ll * (y - (*it));

sx.insert(x);
sy.insert(y);
}

printf("%lld\n", ans );
return 0;
}

inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}