POJ 1724 ROADS

来源：互联网发布：算法入门怎么学编辑：程序博客网时间：2024/04/29 09:47

题目链接：点击打开链接

题目大意：有n个点，给你r条边，再给你k块钱，r行数据中每行4个整数S,D,L,T代表从S到D点的长度为L，需要花费为T，问你从1到n点在k块钱足够的情况下最短路长是多少？

题目解析：简单的搜索，和POJ3411的做法差不多，记得如果当前状态已经大于答案了或者钱已经不够了就要及时return，没有必要再继续搜下去，然后就没啥啦。

#include <algorithm>#include <iostream>#include <numeric>#include <cstring>#include <iomanip>#include <string>#include <vector>#include <cstdio>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#define LL long long#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const int m = 100007;const double esp = 1e-6;const double PI = 3.14159265359;const int INF = 0x3f3f3f3f;using namespace std;const int maxn = 10005;struct node{    int v,len,toll,next;}Map[maxn];int n,r,s,d,l,t,tot;int head[105];int vis[105],ans;void add_edge(){    Map[tot].v = d;    Map[tot].len = l;    Map[tot].toll = t;    Map[tot].next = head[s];    head[s] = tot;    tot += 1;}void dfs(int fr,int toll,int dis){    if(dis > ans)        return ;    if(fr == n && toll >= 0){        ans = min(ans,dis);        return ;    }    for(int i=head[fr];i!=-1;i=Map[i].next){        if(!vis[Map[i].v] && toll >= Map[i].toll){            vis[Map[i].v] = true;            dfs(Map[i].v,toll-Map[i].toll,dis+Map[i].len);            vis[Map[i].v] = false;        }    }    return ;}int main(){    int k;    while(~scanf("%d %d %d",&k,&n,&r)){        memset(Map,INF,sizeof(Map));        memset(head,-1,sizeof(head));        tot = 0;        for(int i=0;i<r;i++){            scanf("%d %d %d %d",&s,&d,&l,&t);            add_edge();        }        memset(vis,false,sizeof(vis));        ans = INF;        dfs(1,k,0);        printf("%d\n",ans==INF?-1:ans);    }    return 0;}

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