POJ - 3616 Milking Time（最长区间调度变形）

最新推荐文章于 2019-08-13 16:55:44 发布

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本文介绍了一种解决最长区间调度问题的算法，通过给定的多个可能重叠的时间区间，目标是在考虑休息限制的情况下，选择一系列区间以最大化产出。文章详细阐述了算法思路，包括区间排序和动态规划的应用。

关于最长区间调度

Milking Time

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
* 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 Nhours

Sample Input

Sample Output

题意：给 m 个区间，现在可以选择若干个区间，要求是两个相邻区间的 right 与 left 的差 >= R. 选择一个区间可以获得相应的值，问最后这个值最大是多少

思路：先对区间按右端点排序（因为二分的时候是根据右端点进行check）。。dp[i] 表示取到 i 个区间可以获得的最大值，那么每一次，可以二分一个符合题意的区间下标 index （有可能没有），那么 dp[i] = max(dp[i - 1],dp[index] + val[i])

Code：

#include<bits/stdc++.h>
#define debug(x) cout << "[" << #x <<": " << (x) <<"]"<< endl
#define pii pair<int,int>
#define clr(a,b) memset((a),b,sizeof(a))
#define rep(i,a,b) for(int i = a;i < b;i ++)
#define pb push_back
#define MP make_pair
#define LL long long
#define ull unsigned LL
#define ls i << 1
#define rs (i << 1) + 1
#define INT(t) int t; scanf("%d",&t)

using namespace std;

const int maxn = 1010;
struct xx{
    int fir,sec;
    int effi;
}interval[maxn];
LL dp[maxn];

bool cmp(xx A,xx B){
    if(A.sec != B.sec)
        return A.sec < B.sec;
    return A.fir < B.fir;
}
int n,m,R;

int Search(int st){
    int l = 0,r = m;
    int ans = 0;
    while(l <= r){
        int mid = (l + r) >> 1;
        if(interval[mid].sec + R > st)
            r = mid - 1;
        else {
            ans = mid;
            l = mid + 1;
        }
    }
    return ans;
}

int main() {
    while(~scanf("%d%d%d",&n,&m,&R)){
        for(int i = 1;i <= m;++ i)
            scanf("%d%d%d",&interval[i].fir,&interval[i].sec,&interval[i].effi);
        sort(interval + 1,interval + 1 + m,cmp);
        for(int i = 1;i <= m;++ i){
            int index = Search(interval[i].fir);
            LL maxx;
            if(index > 0){
                maxx = dp[index] + interval[i].effi;
            }
            else
                maxx = interval[i].effi;
            dp[i] = max(dp[i - 1],maxx);
        }
        printf("%lld\n",dp[m]);
    }
    return 0;
}