CookSteak

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 10), indicating the number of test cases(number of steaks ZJH wants to eat!). For each test case:

The first line contains an integer N (1 ≤ N ≤ $10^5$ ), indicating the number of processes.

Then followed N lines, each line contains two integers $l_i, r_i$ (0 ≤ $l_i$ ≤ $r_i$ ≤ $10^9$ ), indicating the temperature range of the i-th process.

C++11(clang++ 3.9) 解法, 执行用时: 630ms, 内存消耗: 504K, 提交时间: 2020-09-26 16:34:42

#include<bits/stdc++.h>
using namespace std;
int main(){
	int xx,c,d,n,m,i,j;
	long long a,b,x,y;
	cin>>xx;
	for(int l=0;l<xx;l++){
		cin>>n;
		x=0,y=0;
		for(i=0;i<n;i++){
			cin>>a>>b;
			if(x<=a) y=y+a-x+1,x=a;
			else if(x<=b) y=y+1;
			else y=y+x-b+1,x=b;
		}
		cout<<y<<endl;
	}
	return 0;
}

pypy3(pypy3.6.1) 解法, 执行用时: 1695ms, 内存消耗: 29528K, 提交时间: 2020-10-23 17:56:34

t = int(input())
for t in range(0, t):
    n = int(input())
    ans = n
    temp = 0
    for i in range(0, n):
        l, r = map(int, input().split())
        if temp>r:
            ans+=temp-r
            temp = r
        elif temp<l:
            ans+=l-temp
            temp = l
    print(ans)

C++14(g++5.4) 解法, 执行用时: 736ms, 内存消耗: 384K, 提交时间: 2020-09-26 15:53:47

#include<bits/stdc++.h>
using namespace std;

int main(){
	long long t,n;
	cin>>t;
	while(t--){
		cin>>n;
		long long sum=0,c=0,l,r;
		for(long long i=0;i<n;i++){
			cin>>l>>r;
			if(c<l)sum+=(l-c+1),c=l;
			else if(c>r)sum+=(c-r+1),c=r;
			else sum++;
		}
		cout<<sum<<endl;
	}
	return 0;
}

Python3(3.9) 解法, 执行用时: 1938ms, 内存消耗: 2816K, 提交时间: 2021-03-08 14:17:46

for _ in range(int(input())):
    n=int(input())
    t=0
    for i in range(n):
        x,y=map(int,input().split())
        if t>y:
            n=n+t-y
            t=y
        if t<x:
            n=n+x-t
            t=x

    print(n)

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