原题链接 Milking Time Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8764 Accepted: 3647 Description
Bessie is such a hard-working cow. In fact, she is so focused on maximizing her PRoductivity that she decides to schedule her next N (1 ≤ N ≤ 1,000,000) hours (conveniently labeled 0..N-1) so that she produces as much milk as possible.
Farmer John has a list of M (1 ≤ M ≤ 1,000) possibly overlapping intervals in which he is available for milking. Each interval i has a starting hour (0 ≤ starting_houri ≤ N), an ending hour (starting_houri < ending_houri ≤ N), and a corresponding efficiency (1 ≤ efficiencyi ≤ 1,000,000) which indicates how many gallons of milk that he can get out of Bessie in that interval. Farmer John starts and stops milking at the beginning of the starting hour and ending hour, respectively. When being milked, Bessie must be milked through an entire interval.
Even Bessie has her limitations, though. After being milked during any interval, she must rest R (1 ≤ R ≤ N) hours before she can start milking again. Given Farmer Johns list of intervals, determine the maximum amount of milk that Bessie can produce in the N hours.
Input
Line 1: Three space-separated integers: N, M, and RLines 2..M+1: Line i+1 describes FJ’s ith milking interval withthree space-separated integers: starting_houri , ending_houri , and efficiencyiOutput
Line 1: The maximum number of gallons of milk that Bessie can product in the N hoursSample Input
12 4 2 1 2 8 10 12 19 3 6 24 7 10 31 Sample Output
43 Source
USACO 2007 November Silver
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 1e6 + 10;const int maxm = 1e3 + 10;typedef long long ll;typedef struct sec{ int s,e; ll v;}sec;sec a[maxm];ll dp[maxn];bool cmp(sec x,sec y){ return x.e < y.e;}int main(){ int n,m,r; scanf("%d%d%d",&n,&m,&r); for(int i=0;i<m;i++) scanf("%d%d%lld",&a[i].s,&a[i].e,&a[i].v); sort(a,a+m,cmp); for(int i=0;i<m;i++){ int e=a[i].e,s=a[i].s; dp[e]=max(dp[e],dp[e-1]);//顺延的情况 if(s-r<=0) dp[e]=max(dp[e],a[i].v);//由上个可用的时间转移过来 else dp[e]=max(dp[e],dp[s-r]+a[i].v); fill(dp+e,dp+n+1,dp[e]); // for(int j=0;j<=n;j++) cout << dp[j] << " "; // cout << endl; } cout << dp[n] <<endl; return 0;}新闻热点
疑难解答
