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Milking Time
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 R Output * Line 1: The maximum number of gallons of milk that Bessie can product in the N hours Sample Input 12 4 2 1 2 8 10 12 19 3 6 24 7 10 31 Sample Output 43 Source |
思路：先按工作的结束时间从小到大排序，再动态规划。dp[i]表示从头开始取到第i段所获得的最大值。二重循环，如果第i段之前的某个段的结束时间加上r小于等于第i段的开始时间，则更新dp[i]。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
struct St
{
int s,t,w;
}st[maxn];
int dp[maxn];
int n,m,r;
int cmp(St a,St b)
{
return a.t<b.t;
}
int main()
{
int i,j;
while (~sfff(n,m,r))
{
FRL(i,0,m)
sfff(st[i].s,st[i].t,st[i].w);
sort(st,st+m,cmp);
int ans=-1;
FRL(i,0,m)
{
dp[i]=st[i].w;
FRL(j,0,i)
{
if (st[j].t+r<=st[i].s)
dp[i]=max(dp[i],dp[j]+st[i].w);
}
ans=max(ans,dp[i]);
}
pf("%d\n",ans);
}
return 0;
}
原文地址：http://blog.csdn.net/u014422052/article/details/44173693
