【noip模拟题】挖掘机（模拟题+精度）

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mkpii make_pair<int, int>
#define pdi pair<double, int>
#define mkpdi make_pair<double, int>
#define pli pair<ll, int>
#define mkpli make_pair<ll, int>
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }
#define printarr1(a, b) for1(_, 1, b) cout << a[_] << ‘\t‘; cout << endl
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=200005;
int a[N], b[N], n, m, bn, an;
const double eps=1e-5;
char s[10];
int main() {
	read(n); read(m);
	for1(i, 1, n) {
		scanf("%s", s);
		int t=getint();
		if(s[0]==‘T‘) b[++bn]=t;
		else { if(t<m) a[++an]=t; }
	}
	sort(b+1, b+1+bn);
	sort(a+1, a+1+an);
	int i=1, j=1;
	double pos=0, ans=0;
	while(i<=an && j<=bn) {
		double sp=(1.0/(double)(i+j-1));
		double at=((double)a[i]-pos)*(i+j-1)+ans;
		if(at<(double)(b[j])) {
			pos=a[i++];
			ans=at;
		}
		else {
			pos+=sp*((double)b[j]-ans);
			if(pos+eps>m) break;
			ans=(double)b[j++];
		}
	}
	while(i<=an && (pos+eps)<=m) { ans+=((double)a[i]-pos)*(double)(i+j-1); pos=a[i++]; }
	while(j<=bn && (pos+eps)<=m) { pos+=(1.0/(double)(i+j-1))*((double)b[j]-ans); ans=(double)b[j++]; }
	if(pos+eps<=m) ans+=(double)(m-pos)*(double)(i+j-1);
	printf("%lld\n", (ll)(ans+eps));
	return 0;
}