TZOJ 1937 Hie with the Pie(floyd+状压dp)

【TZOJ 1937 Hie with the Pie(floyd+状压dp)】的更多相关文章

TZOJ 1937 Hie with the Pie(floyd+状压dp)

描述 The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately, due to cutbacks, they can afford to hire only one driver to do the deliveries. He will wait for 1 or more (up to 10) orders to be processed…

POJ 3311 Hie with the Pie floyd+状压DP

链接:http://poj.org/problem?id=3311 题意:有N个地点和一个出发点(N<=10),给出全部地点两两之间的距离,问从出发点出发,走遍全部地点再回到出发点的最短距离是多少. 思路:首先用floyd找到全部点之间的最短路.然后用状态压缩,dp数组一定是二维的,假设是一维的话不能保证dp[i]->dp[j]一定是最短的.由于dp[i]记录的"当前位置"不一定是能使dp[j]最小的当前位置.所以dp[i][j]中,i表示的二进制下的当前已经经过的状态,j…

POJ3311 Hie with the Pie 【状压dp/TSP问题】

题目链接:http://poj.org/problem?id=3311 Hie with the Pie Time Limit: 2000MS Memory Limit: 65536K Total Submissions:12225 Accepted: 6441 Description The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortuna…

POJ 3311 Hie with the Pie （状压DP）

题意: 每个点都可以走多次的TSP问题:有n个点(n<=11),从点1出发,经过其他所有点至少1次,并回到原点1,使得路程最短是多少? 思路: 同HDU 5418 VICTOR AND WORLD (可重复走的TSP问题,状压DP)这道题几乎一模一样. //#include <bits/stdc++.h> #include <iostream> #include <cstdio> #include <cstring> #include <cmat…

【POJ3311】Hie with the Pie（状压DP，最短路）

题意: 思路:状压DP入门题 #include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include<iostream> #include<vector> #include<map> #include<set> #define oo 1000000000 #define N (1<<12)-1 using n…

POJ 3311 Hie with the Pie（状压DP + Floyd）

题目链接:http://poj.org/problem?id=3311 Description The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately, due to cutbacks, they can afford to hire only one driver to do the deliveries. He will wait fo…