POJ 1724 ROADS (有限制的最短路徑DFS/BFS）

題意：有n座城市，城市之間有道路，道路需要收費，現在Bob想從城市1去城市n，但是他所擁有的錢是有限制的。現在問Bob能否在有限的錢之內到達n城，若能則輸出最短路徑。
題解：一開始用vector建圖，果斷TLE··。可能是從尾部添加鄰接點的原因吧。

#include <vector>#include <iostream>using namespace std;#define N  10005#define INF 99999struct Edge{int v, len, cost, next;} edge[N];int vis[101], head[N];int ans, k, n, r;void dfs ( int u, int len, int money ){if ( len >= ans )return;if ( u == n ){ans = len;return;}for ( int i = head[u]; i ; i = edge[i].next ){if ( !vis[edge[i].v] &&  money >= edge[i].cost ){vis[edge[i].v] = true;        dfs ( edge[i].v, len + edge[i].len, money-edge[i].cost );    vis[edge[i].v] = false;}}}int main(){int u, v, l, c;memset(vis,0,sizeof(vis));memset(edge,0,sizeof(edge));memset(head,0,sizeof(head));scanf("%d%d%d",&k,&n,&r);int i = 0;while ( r-- ){scanf("%d%d%d%d",&u,&v,&l,&c);i++;edge[i].v = v;edge[i].len = l;edge[i].cost = c;edge[i].next = head[u];head[u] = i;}ans = INF;vis[1] = true;dfs ( 1, 0, k );if ( ans == INF ) printf("-1\n");else printf("%d\n",ans); return 0;}

#include <queue>#include <iostream>using namespace std;#define N  10005struct Edge{int v, len, cost, next;} edge[N];struct Node{int v, len, cost;friend bool operator<(Node a, Node b){        return a.len > b.len;}} node[N];int head[N];int k, n, r;priority_queue<Node> que; /* 優先隊列可以保證每次取出距離最小的點 */int BFS(){Node now, next;now.cost = 0;now.len = 0;now.v = 1;que.push(now);while ( ! que.empty () ){now = que.top ();que.pop();if ( now.v == n )return now.len;for ( int i = head[now.v]; i; i = edge[i].next ){if ( edge[i].cost + now.cost <= k ){next.v = edge[i].v;next.len = edge[i].len + now.len;next.cost = edge[i].cost + now.cost;que.push(next);}}}return -1;}int main(){int u, v, l, c;scanf("%d%d%d",&k,&n,&r);int i = 0;memset(head,0,sizeof(head));while ( r-- ){scanf("%d%d%d%d",&u,&v,&l,&c);i++;edge[i].v = v;edge[i].len = l;edge[i].cost = c;edge[i].next = head[u];head[u] = i;}int ans = BFS();if ( ans == -1 ) printf("-1\n");else printf("%d\n",ans); return 0;}