Purple Book Exercises 8-8 UVa 1612 (Greedy + Precision)

I wrote this question very quickly, but I kept WA, and then I found out that it was precision, which took me an hour...

(1) Greed. Try to score as high as possible each time, you can guarantee the highest score in the end

(2) God tm accuracy problem. Remember to use EPS (1e-6) when judging greater than less than and equal to

a == b                  fabs(a-b) < EPS

a != b                  fabs(a-b) > EPS

a < b                   a + EPS < b

a <= b                   a - EPS < b

a > b                   b + EPS < a

a >= b                   b - EPS < a

#include<cstdio>
#include<vector>
#include<algorithm>
#include<cmath>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;

const int MAXN = 21234;
const double EPS = 1e-6;
vector<double> score[MAXN];
int num[MAXN], n;

void add(int i, double x, double y, double z)
{
	score[i].clear();
	score[i].push_back(0);
	score[i].push_back(x); score[i].push_back(y);
	score[i].push_back(z); score[i].push_back(x+y);
	score[i].push_back(y+z); score[i].push_back(x+z);
	score[i].push_back(x+z+y);
	sort(score[i].begin(), score[i].end());
}

intmain()
{
	int case = 0;
	while(~scanf("%d", &n) && n)
	{
		REP(i, 0, n)
		{
			double x, y, z;
			scanf("%lf%lf%lf", &x, &y, &z);
			add(i, x, y, z);
		}
		
		REP(i, 0, n) scanf("%d", &num[i]), num[i]--;
		double years = 2e9;
		bool ok = true;
		
		REP(i, 0, n)
		{
			int x = num[i], pos = 0;
			while(pos < score[x].size() && score[x][pos] + EPS < ans) pos++;
			if((i == 0 || num[i] > num[i-1]) && pos < score[x].size() && fabs(score[x][pos] - ans ) < EPS) continue;
			if(pos - 1 < 0)  { ok = false; break; }
			ans = score[x][pos-1];
		}
		
		if(ok) printf("Case %d: %.2lf\n", ++kase, ans);
		else printf("Case %d: No solution\n", ++kase);
	}
	
	return 0;
}