POJ 3171 (minimum interval coverage cost)

Language:
Cleaning Shifts
Time Limit: 1000 MS Memory Limit: 65536 K
Total Submissions: 2093 Accepted: 735
Description
Returns the minimum price of covering [M, E] (0 <= M <= E <= 10,000) with N (1 <= N <= 86,399) intervals.
The price of each interval is S (where 0 <= S <= 500,000 ).

Input
3 integers in the first row: N, M, E.

From the second row to the n + 1 row, each row has three numbers, representing the left endpoint T1, the right endpoint T2, and the cost S.
Output www.2cto.com
Only one line indicates the minimum cost, with no solution loss-1.
Sample Input
3 0 4
0 2 3
3 4 2
0 0 1
Sample Output
5
Hint
Example
Take the first and second intervals.
Source
USACO 2005 December Silver
This is a Dp problem. List the Dp equation first.
F [I] indicates the cost of the range [M, I ].
Apparently F [M-1] = 0, the answer is F [E]
Then the equation is F [a [I]. T2] = min (F [j]) + a [I]. S (T1-1 <= J <= T2-1)
A [I] is arranged in ascending order by T2;
Then apparently a [I] To get, [M, T1-1] has been given in front,
If it is obtained by the following [t1, t2], then there must be t1 <T1 T2 <t2, and there is no need to take [T1, T2.

The minimum number can be used as the O (NlogN) by using the line segment tree ).

[Cpp]
# Include <cstdio>
# Include <cstring>
# Include <cstdlib>
# Include <cmath>
# Include <cctype>
# Include <iostream>
# Include <functional>
# Include <algorithm>
Using namespace std;
# Define MAXN (10000 + 10)
# Define MAXE (86399)
# Define MAXS (500000 + 10)
# Define INF (9187201950435737471)
Int n, s, e;
Struct SegMent
{
Int l, r;
Long S;
SegMent (){}
SegMent (int _ l, int _ r, long _ S): l (_ l), r (_ r), S (_ S ){}
Friend bool operator <(const SegMent a, const SegMent B) {return a. r <B. r ;}
} A [MAXN];
Struct SegMentTree // min ()
{
Int n, M;
Long t [MAXE * 10];
Void fillchar (int _ n)
{
N = _ n + 2;
M = 1; while (M-2 <n) M <= 1;
Memset (t, 127, sizeof (t ));
}
Void update (int x)
{
For (x> = 1; x> = 1) t [x] = min (t [x <1], t [(x <1) ^ 1]);
}
Void insert (int x, long c)
{
X = x + 2;
X + = M;
If (t [x]> c) {t [x] = c; update (x );}
}
Long find (int l, int r)
{
L = l + 2; r = r + 2;

L = L-1 + M; r = r + 1 + M;
Long ans = INF;
While (l ^ r ^ 1)
{
If (~ L & 1) ans = min (ans, t [l + 1]);
If (r & 1) ans = min (ans, t [r-1]);
L >>> = 1; r >>> = 1;
}
Return ans;
}
} T;
Int main ()
{
// Freopen ("poj3171.in", "r", stdin );
Scanf ("% d", & n, & s, & e );
For (int I = 1; I <= n; I ++) scanf ("% d", & a [I]. l, & a [I]. r, & a [I]. S );
Sort (a + 1, a + 1 + n );
T. fillchar (e );
T. insert (S-1, 0 );
For (int I = 1; I <= n; I ++)
{
If (a [I]. r <s) continue;
T. insert (a [I]. r, t. find (max (S-1, a [I]. l-1), a [I]. r-1) + a [I]. S );
}
/* For (int I = t. M; I <= t. M * 2; I ++) cout <t. t [I] <'';
Cout <endl;
*/If (t. t [e + 2 + t. M] = INF) cout <"-1 \ n ";
Else cout <t. t [e + 2 + t. M] <endl;


Return 0;
}