Bzoj 1642: [usaco2007 Nov] milking time

Description

Bessie is a very hard-working cow who is always focused on improving her production. To produce more milk, she expected the next N (1 ≤ n ≤ 1,000,000) hours, marked as 0 .. N-1. Farmer John planned M (1 ≤ m ≤ 1,000) milking time periods. Each time period has a start time (0 ≤ start time ≤ n), an end time (start time <End time ≤ n), and a yield (1 ≤ output ≤ 1,000,000) indicates the amount of milk that can be milked from Bessie. Farmer John milking from the start time to the end time respectively. The entire time period must be used for each milking. But even Bessie has her production limit. After each milking, she had to rest r (1 ≤ r ≤ n) hours before the next milking. Given the time range planned by farmer John, calculate the maximum amount of milking in N hours. Input

Three integers n, m, R. Next m in row 1st, three integers Si, EI, Pi. Output in each row

Maximum milk yield.

Question:

First, sort the time period by the end time, and define f [I] as the maximum benefit from the start time to the period I.

F [I] = max {f [J]} + W [I]

J <I and the end time of J + R is before the start time of I.

Ans = max {f [I]}

Code:

#include<cstdio>#include<cstring>#include<algorithm>//by zrt//problem:using namespace std;int n,m,r;int f[1005];struct N{    int s,e,p;    friend bool operator < (N a,N b){        return a.e<b.e;    }}a[1005];int inf=1<<30;int main(){    #ifdef LOCAL    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);    #endif    scanf("%d%d%d",&n,&m,&r);    for(int i=1;i<=m;i++){        scanf("%d%d%d",&a[i].s,&a[i].e,&a[i].p);    }    a[0].e=-inf;    sort(a,a+1+m);    int ans=0;    for(int i=1;i<=m;i++){        for(int j=0;j<i;j++){            if(a[j].e+r<=a[i].s){                f[i]=max(f[i],f[j]+a[i].p);            }        }        ans=max(ans,f[i]);    }         printf("%d\n",ans);    return 0;}

Bzoj 1642: [usaco2007 Nov] milking time