POJ 1040:Transportation

Transportation

Time Limit: 1000MS
Memory Limit: 10000K

Description

Ruratania is just entering capitalism and is establishing new enterprising activities in many fields in- cluding transport. The transportation company TransRuratania is starting a new express train from city A to city B with several stops in the stations on the way. The stations are successively numbered, city A station has number 0, city B station number m. The company runs an experiment in order to improve passenger transportation capacity and thus to increase its earnings. The train has a maximum capacity n passengers. The price of the train ticket is equal to the number of stops (stations) between the starting station and the destination station (including the destination station). Before the train starts its route from the city A, ticket orders are collected from all onroute stations. The ticket order from the station S means all reservations of tickets from S to a fixed destination station. In case the company cannot accept all orders because of the passenger capacity limitations, its rejection policy is that it either completely accept or completely reject single orders from single stations.

Write a program which for the given list of orders from single stations on the way from A to B determines the biggest possible total earning of the TransRuratania company. The earning from one accepted order is the product of the number of passengers included in the order and the price of their train tickets. The total earning is the sum of the earnings from all accepted orders.

Input

The input file is divided into blocks. The first line in each block contains three integers: passenger capacity n of the train, the number of the city B station and the number of ticket orders from all stations. The next lines contain the ticket orders. Each ticket order consists of three integers: starting station, destination station, number of passengers. In one block there can be maximum 22 orders. The number of the city B station will be at most 7. The block where all three numbers in the first line are equal to zero denotes the end of the input file.

Output

The output file consists of lines corresponding to the blocks of the input file except the terminating block. Each such line contains the biggest possible total earning.

Sample Input

10 3 4
0 2 1
1 3 5
1 2 7
2 3 10
10 5 4
3 5 10
2 4 9
0 2 5
2 5 8
0 0 0

Sample Output

19
34

思路：

先按照站点的顺序对票的序列进行预处理，之后递归逐个搜索按照加还是不加的顺序搜索票的序列，找到最大值即可

AC代码：

#include
#include
#include
#include

using namespace std;
int n, staNum, ticNum;

struct node
{
	int s;
	int e;
	int p;
	node(){}
	bool operator < (const node b)const {
		if (s != b.s)
			return s < b.s;
		else
			return e < b.e;
	}
}tick[23];
int sum = 0, off[9];
int ans = 0;

void dfs(int cnt, int tot) {
	if (cnt >= ticNum) {
		ans = max(sum, ans);
		return;
	}
	if (cnt > 0) {
		for (int i = tick[cnt - 1].s + 1; i <= tick[cnt].s; ++i)
			tot -= off[i];
	}
	tot += tick[cnt].p;
	if (tot <= n) {
		off[tick[cnt].e] += tick[cnt].p;
		sum += (tick[cnt].e - tick[cnt].s) * tick[cnt].p;
		dfs(cnt + 1, tot);
		sum -= (tick[cnt].e - tick[cnt].s) * tick[cnt].p;
		off[tick[cnt].e] -= tick[cnt].p;
	}
	tot -= tick[cnt].p;
	dfs(cnt + 1, tot);
}

int main() 
{
	int i;
	while (scanf("%d%d%d", &n, &staNum, &ticNum), n != 0 || staNum != 0 || ticNum != 0) {
		memset(off, 0, sizeof off);
		ans = 0; sum = 0;
		for (i = 0; i < ticNum; ++i) {
			scanf("%d%d%d", &tick[i].s, &tick[i].e, &tick[i].p);
		}
		sort(tick, tick + ticNum);
		dfs(0, 0);
		printf("%d\n", ans);
	}
	return 0;
}