R - Milking Time——POJ C语言实现

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R - Milking Time

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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_hour_i ≤ N), an ending hour (starting_hour_i < ending_hour_i ≤ N), and a corresponding efficiency (1 ≤ efficiency_i ≤ 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_hour_i , ending_hour_i , and efficiency_i

Output

* Line 1: The maximum number of gallons of milk that Bessie can product in the N hours

Sample Input

12 4 21 2 810 12 193 6 247 10 31

Sample Output

C语言实现，先对时间段进行排序，然后DP代表前N段的最大值

#include<stdio.h>#include<string.h>int my_max(int x,int y)   {return x>y?x:y;}typedef struct node{int star;int end;int eff;}NODE;  NODE que[1005];int dp[1005];int main(){int n,m,r,i,j,max;NODE t;while(scanf("%d %d %d",&n,&m,&r)!=EOF){memset(que,0,sizeof(que));memset(dp,0,sizeof(dp));for(i=0;i<m;i++)  scanf("%d %d %d",&que[i].star,&que[i].end,&que[i].eff);for(i=0;i<m;i++){for(j=i;j<m;j++){if(que[i].star>que[j].star){t=que[j];que[j]=que[i];que[i]=t;}}}for(i=0;i<m;i++){max=0;for(j=0;j<i;j++){if(que[i].star-que[j].end>=r){max=my_max(max,dp[j]);}} dp[i]=max+que[i].eff;}max=0; for(j=0;j<m;j++)//求出最大值 if(max<dp[j])max=dp[j];         printf("%d\n",max); }return 0;}

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=10005;struct node{    int l,r,data;    bool operator<(const node& tmp) const{        if(l!=tmp.l)            return l<tmp.l;//如果区间开始不一样，则按开始为标准比较         else            return r<tmp.r;//如果区间开始一样，则按结束为标准比较     }}p[maxn];int dp[maxn];//表示选择到第M个区间 能获得的最大值int main(){    int n,m,r;    while(~scanf("%d%d%d",&n,&m,&r)){        for(int i=0;i<m;i++){            scanf("%d%d%d",&p[i].l,&p[i].r,&p[i].data);            dp[i]=0;//初始化DP         }        sort(p,p+m);//排序         int j;        for(int i=0;i<m;i++){            int maxx=0;            for(j=0;j<i;j++)if(p[i].l-p[j].r>=r)//有一段r的距离 {               maxx=max(maxx,dp[j]);//找到这个区间前面的最大值           }            dp[i]=maxx+p[i].data;        }        int max=0;        for(j=0;j<m;j++)//求出最大值 if(max<dp[j])max=dp[j];         printf("%d\n",max);    }    return 0;}

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