Codeforces #186(div 2)D. Ilya and Roads

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 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 %I64dspecifier.

Sample test(s)

input

10 4 67 9 116 9 137 7 73 5 6

output

17

input

10 7 13 4 158 9 85 6 89 10 61 4 21 4 108 10 13

output

2

input

10 1 95 10 14

output

-1

题目大意:给定n个公司修理某个区间的路需要的钱数，求修好n单位的路最少需要多少钱。

大致思想:纯dp.用dp[i][j]表示修好前i单位长度中j单位长度的路最少需要多少钱,则

dp[i][j]=min(dp[i][j],dp[k][j-i+k]+num[k][i]) 0<k<i

其中num[i][j]表示i到j段被一个公司修好最少需要多少钱。

虽然看起来很容易，但是能想把这些想明白，也是挺困难的……

难度：4 个人感觉比上一道dp难一些，可能是因为状态多了吧。dp还需要加强呀。

#include<bits/stdc++.h>using namespace std;#define ll long longll dp[303][303];ll num[303][303];const ll inf=0x123456789ABCDE;int main(){ll m,n,i,j,k,a,b,c,ans=inf,s;for(i=0;i<=302;i++){fill(num[i],num[i]+302,inf);fill(dp[i],dp[i]+302,inf);}dp[0][0]=0ll;cin>>n>>m>>s;while(m--){cin>>a>>b>>c;num[a-1][b]=min(num[a-1][b],c);}    for(i=1;i<=n;i++){   for(j=n;j>=i;j--){             num[i+1][j]=min(num[i+1][j], num[i][j]);             num[i][j-1]=min(num[i][j-1], num[i][j]);        }    }/*for(i=0;i<=n;i++){for(j=0;j<=n;j++) cout<<num[i][j]<<' ';cout<<endl;}*/for(i=1;i<=n;i++){for(j=0;j<=i;j++){dp[i][j]=dp[i-1][j];for(k=0;k<i;k++){dp[i][j]=min(dp[i][j],dp[k][j-i+k]+num[k][i]);}}}/*for(i=1;i<=n;i++){for(j=1;j<=n;j++) cout<<dp[i][j]<<' ';cout<<endl;}*/for(i=n;i>=s;i--) ans=min(ans,dp[n][i]);cout<<(ans==inf?-1:ans)<<endl; }