POJ 3616 Milking Time

Milking Time

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3616
64-bit integer IO format: %lld      Java class name: Main

 

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 N hours

 

Sample Input

12 4 2
1 2 8
10 12 19
3 6 24
7 10 31

Sample Output

43

Source

USACO 2007 November Silver

 

解题：观察发现，M比较小，故想到转换成图论，求最长路来解题。。。要注意数据比较大，用long long才过瘾

 

/*
@author: Lev
@date:
*/
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <climits>
#include <deque>
#include <sstream>
#include <fstream>
#include <bitset>
#include <iomanip>
#define LL long long
#define INF 0x3f3f3f3f

using namespace std;
const int maxn = 1000010;
struct arc{
    int to,next,cost;
    arc(int x = 0,int y = 0,int z = -1){
        to = x;
        cost = y;
        next = z;
    }
};
arc e[maxn];
int head[maxn],tot,x[maxn],y[maxn],c[maxn];
LL d[maxn];
void add(int u,int v,int cost){
    e[tot] = arc(v,cost,head[u]);
    head[u] = tot++;
}
queue<int>q;
bool in[maxn];
void spfa(){
    for(int i = 0; i < maxn; ++i){
        d[i] = 0;
        in[i] = false;
    }
    while(!q.empty()) q.pop();
    q.push(0);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        in[u] = false;
        for(int i = head[u]; ~i; i = e[i].next){
            if(d[e[i].to] < d[u] + e[i].cost){
                d[e[i].to] = d[u] + e[i].cost;
                if(!in[e[i].to]){
                    in[e[i].to] = true;
                    q.push(e[i].to);
                }
            }
        }
    }
}
int main(){
    int N,M,R;
    while(~scanf("%d %d %d",&N,&M,&R)){
        for(int i = 1; i <= M; ++i)
            scanf("%d %d %d",x+i,y+i,c+i);
        memset(head,-1,sizeof(head));
        tot = 0;
        for(int i = 1; i <= M; ++i)
        for(int j = 1; j <= M; ++j)
            if(x[j] >= y[i]+R) add(i,j,c[j]);
        for(int i = 1; i <= M; ++i)
            add(0,i,c[i]);
        spfa();
        LL ans = 0;
        for(int i = 1; i <= M; ++i)
            ans = max(ans,d[i]);
        printf("%I64d\n",ans);
    }
    return 0;
}