Codeforces Round #186 (Div. 2)---D. Ilya and Roads

Everything is great about Ilya’s city, except the roads. The thing is, the only ZooVille road is represented as n holes in a row. We will consider the holes numbered from 1 to n, from left to right.

Ilya is really keep on helping his city. So, he wants to fix at least k holes (perharps he can fix more) on a single ZooVille road.

The city has m building companies, the i-th company needs ci money units to fix a road segment containing holes with numbers of at least li and at most ri. The companies in ZooVille are very greedy, so, if they fix a segment containing some already fixed holes, they do not decrease the price for fixing the segment.

Determine the minimum money Ilya will need to fix at least k holes.
Input

The first line contains three integers n, m, k (1 ≤ n ≤ 300, 1 ≤ m ≤ 105, 1 ≤ k ≤ n). The next m lines contain the companies’ description. The i-th line contains three integers li, ri, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ci ≤ 109).
Output

Print a single integer — the minimum money Ilya needs to fix at least k holes.

If it is impossible to fix at least k holes, print -1.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Sample test(s)
Input

10 4 6
7 9 11
6 9 13
7 7 7
3 5 6

Output

17

Input

10 7 1
3 4 15
8 9 8
5 6 8
9 10 6
1 4 2
1 4 10
8 10 13

Output

2

Input

10 1 9
5 10 14

Output

-1

dp[i][j] 表示前i个点，埋了j个的最小代价
C[i][j]表示从i到j雇佣一家公司埋好花的最少的钱

/************************************************************************* > File Name: CF-186-D.cpp > Author: ALex > Mail: [email protected] > Created Time: 2015年04月02日 星期四 16时39分36秒 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const double eps = 1e-15;
typedef long long LL;
const LL inf = (1LL << 60);
typedef pair <int, int> PLL;

const int N = 330;
LL C[N][N];
LL dp[N][N];

int main()
{
    int n, m, k;
    while (~scanf("%d%d%d", &n, &m, &k))
    {
        int l, r;
        LL c;
        for (int i = 0; i <= n + 1; ++i)
        {
            for (int j = 0; j <= n + 1; ++j)
            {
                dp[i][j] = inf;
                C[i][j] = inf;
            }
        }
        for (int i = 0; i <= n; ++i)
        {
            dp[i][0] = 0;
        }
        for (int i = 1; i <= m; ++i)
        {
            scanf("%d%d%lld", &l, &r, &c);
            C[l][r] = min(C[l][r], c);
        }
        for (int i = 1; i <= n; ++i)
        {
            for (int j = n; j >= i; --j)
            {
                C[i][j] = min(C[i][j], C[i - 1][j]);
                C[i][j] = min(C[i][j], C[i][j + 1]);
            }
        }
        for (int i = 1; i <= n; ++i)
        {
            for (int j = 1; j <= i; ++j)
            {
                dp[i][j] = dp[i - 1][j];
                for (int p = 0; p < i; ++p)
                {
                    if (C[p + 1][i] == inf)
                    {
                        continue;
                    }
                    dp[i][j] = min(dp[i][j], dp[p][j - i + p] + C[p + 1][i]);
                }
            }
        }
        LL ans = inf;
        for (int i = 1; i <= n; ++i)
        {
            for (int j = k; j <= i; ++j)
            {
                ans = min(ans, dp[i][j]);
            }
        }
        if (ans >= inf)
        {
            ans = -1;
        }
        printf("%lld\n", ans);
    }
    return 0;
}