题目大意，N个区间覆盖[T1,T2]及对应的代价S，求从区间M到E的全部覆盖的最小代价是多少。 (1 <= N <= 10,000)，(0 <= M <= E <= 86,399).

      思路是DP,首先将每个区间按照T2从小到大排序，设dp(k)为从m覆盖到k所需最小代价，则有

dp(T2[i]) = min(dp(T2[i]), {dp(j) + Si,  T1[i] - 1<=j <= T2[i]}),对于 {dp(j) + Si,  T1[i] - 1<=j <= T2[i]}我们可以用线段树来进行优化，所以最终复杂度为O(n*logE)。

#include <stdio.h>
#include <vector>
#include <math.h>
#include <string.h>
#include <string>
#include <iostream>
#include <queue>
#include <list>
#include <algorithm>
#include <stack>
#include <map>
#include <time.h>
using namespace std;

struct T
{
	int t1;
	int t2;
	int S;
};
#define MAXV 5000000001
long long BinTree[270000];
T cows[10001];
long long DP[86401];

template<class TYPE>
void UpdateValue(TYPE st[],int i, TYPE value, int N, bool bMin)
{
	i += N - 1;
	st[i] = value;
	while (i > 0)
	{
		i = (i - 1) / 2;
		if (bMin)
		{
			st[i] = min(st[i * 2 + 1], st[i * 2 + 2]);
		}
		else
			st[i] = max(st[i * 2 + 1], st[i * 2 + 2]);
	}
}

template<class TYPE>
TYPE QueryST(TYPE st[], int a, int b, int l, int r, int k, bool bMin)
{
	if (l > b || a > r)
	{
		return bMin ? MAXV : 0;
	}
	if (l >= a && b >= r)
	{
		return st[k];
	}
	else
	{
		TYPE value1 = QueryST(st, a, b, l, (r + l) / 2, k * 2 + 1, bMin);
		TYPE value2 = QueryST(st, a, b, (r + l) / 2 + 1, r, k * 2 + 2, bMin);
		if (bMin)
		{
			return min(value1, value2);
		}
		else
		{
			return max(value1, value2);
		}
	}
}

int compT(const void* a1, const void* a2)
{
	if (((T*)a1)->t2 - ((T*)a2)->t2 == 0)
	{
		return ((T*)a1)->t1 - ((T*)a2)->t1;
	}
	else
		return ((T*)a1)->t2 - ((T*)a2)->t2;
}


int main()
{
#ifdef _DEBUG
	freopen("e:\\in.txt", "r", stdin);
#endif
	int N, M, E;
	scanf("%d %d %d", &N, &M, &E);
	M++;
	E++;
	for (int i = 0; i < N; i++)
	{
		scanf("%d %d %d", &cows[i].t1, &cows[i].t2, &cows[i].S);
		cows[i].t1++;
		cows[i].t2++;
	}
	int maxe = 1;
	while (maxe < E)
	{
		maxe *= 2;
	}
	for (int i = 0; i < maxe * 2;i++)
	{
		BinTree[i] = MAXV;
	}
	for (int i = 0; i <= E;i++)
	{
		DP[i] = MAXV;
	}

	DP[M - 1] = 0;
	UpdateValue<long long>(BinTree, M - 1, 0, maxe, true);
	qsort(cows, N, sizeof(T), compT);
	for (int i = 0; i < N;i++)
	{
		DP[cows[i].t2] = min(DP[cows[i].t2], QueryST<long long>(BinTree, cows[i].t1 - 1, cows[i].t2, 0, maxe - 1, 0, true) + cows[i].S);
		UpdateValue<long long>(BinTree, cows[i].t2, DP[cows[i].t2], maxe, true);
	}
	if (E <= cows[N - 1].t2)
	{
		DP[E] = QueryST<long long>(BinTree, E, cows[N - 1].t2, 0, maxe - 1, 0, true);
	}
	
	if (DP[E] >= MAXV)
	{
		printf("-1\n");
	}
	else
		printf("%I64d\n", DP[E]);
	return 0;
}