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Language: Default 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), 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 Input * Line 1: Three space-separated integers: N, M, and R * Lines 2..M+1: Line i+1 describes FJ‘s ith milking interval withthree space-separated integers: starting_houri , ending_houri , and efficiencyi 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 |
题意：给个时间长度n，m个工作时间段和每个时间段能完成的工作量，一次只能做一个工作并且一旦开始做就要把它做完，要求选择的两个工作时间段之间至少相差r时间（中间需要休息嘛）求选择那些工作n时间内能完成的最大工作量。输出最大值。
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;
}