題意：

N個城市，編號1到N。城市間有R條單向道路。有長度和過路費兩個屬性。Bob只有K塊錢，他想從城市1走到城市N。問最短共需要走多長的路。如果到不了N，輸出-1。

題目：

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

分析：

給了n個城市，需要從城市1到城市n,求最短距離，要是這樣就是一道簡單的最短路問題，但題目有限制要求，在滿足花費即不超過k的情況下的最短路
1.首先考慮得用搜索，迭代尋找，在找到所有滿足路徑后，選出最短的（注意剪枝）。
2.選用鄰接表來存儲無向圖的時間空間復雜度是O(M)。
如果只是查詢兩個點之間是否相鄰,鄰接矩陣當然更快,但如果是做dfs的話,找當前節點相鄰的點,如果用鄰接矩陣的話每次都要從1掃到n,如果用鄰接表遍歷每個頂點的邊

AC代碼

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; typedef long long ll; const int inf=0x3f3f3f3f; const int M=1e4+10; int n,m,k; ll ans; int a[M],b[M],c[M],d[M],fir[M],nex[M],book[M]; void dfs(int x/*當前到達城市*/,int y/*費用*/,int z/*路程*/) {if(y>n||z>=ans)return ;if(x==m){ans=z;return ;}for(int i=fir[x];i;i=nex[i])if(!book[i]){book[i]=1;dfs(b[i],y+d[i],z+c[i]);book[i]=0;}return ; } int main() {scanf("%d%d%d",&n,&m,&k);for(int i=1;i<=k;i++)/*care 變量要從1開始，鄰接表定義*/{scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);nex[i]=fir[a[i]];fir[a[i]]=i;}ans=inf;dfs(1,0,0);if(ans==inf) printf("-1\n");else printf("%lld\n",ans);return 0; }

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