Codeforces 313D Ilya and Roads【Dp+思维】

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针对一条包含多个洞的道路，本文介绍了一种算法来确定修复至少K个洞所需的最小花费。通过动态规划的方法，预处理出修复任意两个洞之间的最小成本，并利用这些信息有效地解决了问题。

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 c_i money units to fix a road segment containing holes with numbers of at least l_i and at most r_i. 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 l_i, r_i, c_i (1 ≤ l_i ≤ r_i ≤ n, 1 ≤ c_i ≤ 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.

Examples

Input

Output

Input

Output

Input

10 1 9
5 10 14

Output

-1

题目大意：

给你N个洞，需要进行填补至少K个，现在有M个施工队，对于这M个施工队的元素有三个：

L,R,C.表示这个施工队只能对于区间【L,R】的洞进行填补，雇佣这个施工队就需要花费C单位的价钱。

问至少填补K个洞的最小花费。

思路：

显然直接贪心是不可取的，我们考虑Dp.对于Dp来讲，对于最终结果的无非有三种状态：

①位子

②填补了多少个洞

③花费为多少。

那么我们设定Dp【i】【j】表示为从左到右的Dp，到位子i，一共填补了j个洞口的最小花费。

那么状态转移过程我们可以O（n*k*m）去求，然而时间复杂度不允许；

所以我们要优化这个m.观察到数据范围N不大，所以我们考虑预处理出cost【i】【j】表示填补区间【i，j】的洞口的最小花费.

因为是线性的动态规划，所以预处理cost【i】【j】的时候没有必要O（m*n^2）去做，直接O（m*n*2）的以左右端点各作为起点和终点去画一段即可：

那么状态转移方程就可以O（n^3）的去转移：

注意实现代码的细节就没有别的难点了。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define ll __int64
struct node
{
    int x,y;
    ll w;
}a[100040];
ll cost[305][305];
ll dp[305][305];
const ll INF=100000000000000000;
int main()
{
    int n,m,kk;
    while(~scanf("%d%d%d",&n,&m,&kk))
    {
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                cost[i][j]=INF;
                dp[i][j]=INF;
            }
        }
        for(int i=1;i<=m;i++)
        {
            int x,y,w;
            scanf("%d%d%I64d",&a[i].x,&a[i].y,&a[i].w);
            for(int j=a[i].x;j<=a[i].y;j++)
            {
                cost[j][a[i].y]=min(cost[j][a[i].y],a[i].w);
                cost[a[i].x][j]=min(cost[a[i].x][j],a[i].w);
            }
            cost[a[i].x][a[i].y]=min(cost[a[i].x][a[i].y],a[i].w);
        }
        dp[0][0]=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=i;j++)
            {
                dp[i][j]=min(dp[i][j],dp[i-1][j]);
                for(int k=1;k<=i;k++)
                {
                    if(j-(i-k+1)>=0)
                    {
                        dp[i][j]=min(dp[i][j],cost[k][i]+dp[k-1][j-(i-k+1)]);
                    }
                }
            }
        }
        if(dp[n][kk]==INF)printf("-1\n");
        else
        printf("%I64d\n",dp[n][kk]);
    }
}