Codeforces Round #186 (Div. 2).D

纠结的一道dp。

状态转移方程还是比较好想的，优化比较纠结

　　　　　　　　　　　　　　　　　　　　　　　　　　D. Ilya and Roads

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 %I64dspecifier.

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

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <map>
#include <queue>
#include <sstream>
#include <iostream>
using namespace std;
#define INF 0x3fffffffffff

typedef __int64 LL;

struct node
{
    int x,y,w;
}g[100100];

int n,m,k;
LL dp[303][303];
LL get[303][303];
LL mx[303];

int cmp(node t,node t1)
{
    if(t.x!=t1.x)
        return t.x<t1.x;
    return t.y>t1.y;
}

//真的如此碉炸天

int main()
{
     //freopen("//home//chen//Desktop//ACM//in.text","r",stdin);
    //freopen("//home//chen//Desktop//ACM//out.text","w",stdout);
    scanf("%d%d%d",&n,&m,&k);
    for(int i=0;i<303;i++)
        for(int j=0;j<303;j++)
            dp[i][j]=INF;
    memset(get,0,sizeof(get));
    memset(g,0,sizeof(g));
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d",&g[i].x,&g[i].y,&g[i].w);
        g[i].y=g[i].y-g[i].x;
    }
    sort(g,g+m,cmp);
    int i=1,j=0;
    LL mi = INF;
    int tmp;
    while( i <= n )
    {
        mi=INF;
        tmp=n;
        while(g[j].x == i)
        {
            for(int i1=tmp;i1>=g[j].y;i1--)
                get[i][i1]=mi;
            tmp=g[j].y;
            mi=min(mi,(__int64)g[j].w);
            j++;
        }
        for(int i1=tmp;i1>=0;i1--)
            get[i][i1]=mi;
        i++;
    }
    for(i=1;i<=k;i++)
        mx[i]=INF; // 0个的时候，不需要w
    // 很好的dp压缩！
    for(i=1;i<=n;i++)
    {
        if(get[i][0]!=INF)
        {
            for(j=0;j <k;j++)
            {
                for(int i1=0;i1+j<k;i1++)
                    dp[i+j][i1+j+1]=min(dp[i+j][i1+j+1],mx[i1]+get[i][j]); //纠结的状态转移方程！
            }
        }
        for(j=0;j<=k;j++)
        {
            mx[j]=min(mx[j],dp[i][j]); // 每一个都是独立的！
            dp[i][j]=mx[j];
        }
    }
    if(mx[k]==INF)
        printf("-1");
    else
    printf("%I64d",mx[k]);
    return 0;
}