摘要: http://poj.org/problem?id=2449题意:给出图,给出起点与终点以及k求第k短路。模板:#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <set>#include <map>#include <str 阅读全文
posted @ 2013-01-22 21:04 E_star 阅读(214) 评论(0) 推荐(0) 编辑
摘要: http://poj.org/problem?id=3255题意:给你一张图,起点为1,终点为n 。求从1到n的次短路。(这里每条边可以重复走)思路:两遍最短路算法分别求出1到所有点的最短距离,n到所有点的最短距离。然后枚举每一条边找出第二短路。#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include &l 阅读全文
posted @ 2013-01-22 16:29 E_star 阅读(409) 评论(0) 推荐(0) 编辑
摘要: 250: (x/k)*3 + (x/3) = p 给出P和K,求满足的x最小的值,直接从1到3*P枚举即可 500:给定两个正数 N,M1 <= A <= N, 1 <= B <= M , SSR(A, B) = (sqrt(A)+sqrt(B))^2 求满足SSR(A, B)为整数的可能数;化简后可得SSR(A, B) = A + B + 2*sqrt(A)*sqrt(B);只要满足sqrt(A)*sqrt(B)为整数即可总的来说就两种情况:1:A和B都是可开方的数;2:都不是可开方的数,但是化简后的根号里面的数相同: sqrt(8) = 2*sqrt(2) 和 sq 阅读全文
posted @ 2013-01-22 11:46 E_star 阅读(308) 评论(2) 推荐(0) 编辑