Java(javac 1.8) 解法, 执行用时: 553ms, 内存消耗: 54548K, 提交时间: 2020-04-06 11:59:14

import java.util.Arrays;
import java.util.Scanner;
 
public class Main
{
    public static void main(String[] args)
    {
        Scanner in=new Scanner (System.in);
        while (in.hasNext())
        {
        	int n=in.nextInt();
        	double t[]=new double[111111];
        	double d[]=new double[111111];
        	int len1=0,len2=0;
        	for (int i=1;i<=n;i++)
        	{
        	    String m=in.next();
        	    double a=in.nextDouble();
        	    if (m.equals("T"))
        	    {
        	    	t[++len1]=a;
        	    }
        	    else
        	    {
        	    	d[++len2]=a;
        	    }
        	}
        	d[++len2]=0;
        	d[++len2]=1000;
        	Arrays.sort(t,1,len1+1);
        	Arrays.sort(d,1,len2+1);
        	double sum=0;
        	int t1=1,t2=1;
        	for (int i=1;i<len2;i++)
        	{
        		double jl=d[i];
        		while(jl<d[i+1] && t1<=len1 && sum+(d[i+1]-jl)*t2>t[t1]  )
        		{
        			jl+=(t[t1]-sum)/t2;
        			t2++;
        			sum=t[t1++];
        		}
        		sum+=(d[i+1]-jl)*t2;
        		t2++;
        	}
        	if (sum-(int)sum>0.5)sum=(int)sum+1;
        	else sum=(int)sum;
        	System.out.printf("%.0f\n",sum);
        } 
    }
}

C++14(g++5.4) 解法, 执行用时: 7ms, 内存消耗: 544K, 提交时间: 2020-04-05 15:52:57

#pragma GCC optimize(3,"Ofast")
#include <bits/stdc++.h>
using namespace std;
template<class t> inline t read(t &x){
	char c=getchar();bool f=0;x=0;
	while(!isdigit(c)) f|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	if(f) x=-x;return x;
}
template<class t> inline void write(t x){
	if(x<0) putchar('-'),write(-x);
	else{if(x>9) write(x/10);putchar('0'+x%10);}
}

const int N=1e5+5;
int x=1,a[N],b[N],nm,n,m;
double y,z;



signed main(){
	read(nm);
	for(int i=1;i<=nm;i++){
		char op;
		cin>>op;
		if(op=='T') read(a[++n]);
		else read(b[++m]);
	}
	sort(a+1,a+1+n);
	sort(b+1,b+1+m);
	for(int l=1,r=1,i=1;i<=nm;i++){
		if((r>m||(b[r]-y)*x+z>=a[l])&&l<=n){
			if((a[l]-z)/x+y>1000) break;
			y+=(a[l]-z)/x;
			z=a[l];
			x++;l++;
		}
		else{
			if(b[r]>1000) break;
			z+=(b[r]-y)*x;
			y=b[r];
			x++;r++;
		}
	}
	write((int)round((1000-y)*x+z)); 
}

C++11(clang++ 3.9) 解法, 执行用时: 11ms, 内存消耗: 492K, 提交时间: 2020-04-04 20:11:09

#include<bits/stdc++.h>
using namespace std;
vector<int>a,b;
int n,num1=0,num2=0;
double ans=0,now=0;
int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		char c;double x;
		cin>>c>>x;
		if(c=='T')a.push_back(x);
		else b.push_back(x);
	}
	sort(a.begin(),a.end());
	sort(b.begin(),b.end());
	a.push_back(1e8);
	b.push_back(1e8);
	for(int i=1;i<=n;i++){
		double v=1.0/i;
		if(b[num2]<(a[num1]-ans)*v+now){
			ans+=(b[num2]-now)/v;
			now=b[num2];
			num2++;
		}
		else{
			now+=(a[num1]-ans)*v;
			ans=a[num1];
			num1++;
		}
	}
	ans+=(1000-now)*(n+1);
	printf("%.0lf",ans);
	return 0;
}