poj 3411 Paid Roads很水的DFS

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poj 3411 Paid Roads很水的DFS

题意:给你N 城市和M条道路,每条道路要付的钱,但是如果你在这个道路上你可以付其他道路的钱(跟走到的时候去的话不一样),问你从1走到N最少话费是多少. 直接DFS搜. 链接http://poj.org/problem?id=3411 代码: #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <stdlib.h> #inc…

多次访问节点的DFS POJ 3411 Paid Roads

POJ 3411 Paid Roads Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6553 Accepted: 2430 Description A network of m roads connects N cities (numbered from 1 to N). There may be more than one road connecting one city with another. Some o…

poj 3411 Paid Roads（dfs）

Description A network of m roads connects N cities (numbered to N). There may be more than one road connecting one city with another. Some of the roads are paid. There are two ways to pay for travel on a paid road i from city ai to city bi: in advanc…

POJ 3411 Paid Roads(DFS)

题目链接点和边都很少,确定一个界限,爆搜即可.判断点到达注意一下,如果之前已经到了,就不用回溯了,如果之前没到过,要回溯. #include <cstring> #include <cstdio> #include <string> #include <iostream> #include <algorithm> #include <vector> #include <queue> using namespace st…

poj 1724 ROADS 很水的dfs

题意:给你N个城市和M条路和K块钱,每条路有话费,问你从1走到N的在K块钱内所能走的最短距离是多少链接:http://poj.org/problem?id=1724 直接dfs搜一遍就是代码: #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <stdlib.h> #include <vector> #i…

POJ 3411 Paid Roads(SPFA || DFS)

题目链接题意 : 要从1城市到n城市,求最短路是多少,从a城市到达b城市的路程,如果你到过c城市,则需要走p,否则走r长. 思路 : 因为可以来回走,所以不能用单纯的最短路,可以用二维SPFA,状态压缩一下,第二维来记录状态,表示到过这个点的第几个状态.也可以用DFS,因为最多十个点,所以如果走某一个点走过三遍说明就是真的只增加费用了,也就是真正的在走环路了.DFS分析二维SPFA #include <stdio.h> #include <string.h> #include…