[POJ 1724] ROADS【DFS】

  • 2018-01-01
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Problem:

Time Limit: 1000MS Memory Limit: 65536K

Description

N cities named with numbers 1 ... N are connected with one-way roads. Each road has two parameters associated with it : the road length and the toll that needs to be paid for the road (expressed in the number of coins).
Bob and Alice used to live in the city 1. After noticing that Alice was cheating in the card game they liked to play, Bob broke up with her and decided to move away - to the city N. He wants to get there as quickly as possible, but he is short on cash.We want to help Bob to find the shortest path from the city 1 to the city N that he can afford with the amount of money he has.

Input

The first line of the input contains the integer K, 0 <= K <= 10000, maximum number of coins that Bob can spend on his way.
The second line contains the integer N, 2 <= N <= 100, the total number of cities.The third line contains the integer R, 1 <= R <= 10000, the total number of roads.

Each of the following R lines describes one road by specifying integers S, D, L and T separated by single blank characters :

  • S is the source city, 1 <= S <= N
  • D is the destination city, 1 <= D <= N
  • L is the road length, 1 <= L <= 100
  • T is the toll (expressed in the number of coins), 0 <= T <=100

Notice that different roads may have the same source and destination cities.

Output

The first and the only line of the output should contain the total length of the shortest path from the city 1 to the city N whose total toll is less than or equal K coins.
If such path does not exist, only number -1 should be written to the output.

Sample Input

5
6
7
1 2 2 3
2 4 3 3
3 4 2 4
1 3 4 1
4 6 2 1
3 5 2 0
5 4 3 2

Sample Output

11

Source

CEOI 1998

Solution:

本题是带有限制的最短路问题,可以使用 DFS 的方法。
DFS 在有向无环图 (Directed Acyclic Graph, DAG) 上进行。
一个小小的最优性剪枝:当前长度已经超过最短路时直接退出。

Code: O(玄学) [320K, 79MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

int K, N, R, minDis;

struct Edge{
	int np, l, t;
	Edge *nxt;
};

struct Graph{
	Edge *V[105], E[10005];
	int tope;
	
	inline void clear() {tope = 0, memset(V, 0, sizeof(V));}
	
	inline void addEdge(int S, int D, int L, int T){
		E[++tope].np = D, E[tope].l = L, E[tope].t = T;
		E[tope].nxt = V[S], V[S] = &E[tope];
	}
} G;

bool vis[105];

inline void DFS(int u, int cur_l, int cur_t){
	if(u == N){
		minDis = min(minDis, cur_l);
		return;
	}
	if(cur_l >= minDis) return;  // Optimality pruning 
	for(register Edge *ne = G.V[u]; ne; ne = ne->nxt){
		if(vis[ne->np] || cur_t + ne->t > K) continue;
		vis[ne->np] = 1;
		DFS(ne->np, cur_l + ne->l, cur_t + ne->t);
		vis[ne->np] = 0;
	}
}

int main(){
	while(scanf("%d", &K) != EOF){
		scanf("%d%d", &N, &R);
		G.clear();
		for(register int i = 1; i <= R; i++){
			int S, D, L, T;
			scanf("%d%d%d%d", &S, &D, &L, &T);
			G.addEdge(S, D, L, T);
		}
		memset(vis, 0, sizeof(vis)), minDis = 0x3f3f3f3f;
		vis[1] = 1, DFS(1, 0, 0);
		if(minDis == 0x3f3f3f3f) puts("-1");
		else printf("%d\n", minDis);
	}
	return 0;
}

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OPEN AT 2017.12.10

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