题意: 给定n个点, m条路, 求1到 2 ~n的最短路之和加上2~n到1的最短路之和 分析: 裸最短路, 求其他点到源点的距离只需要把边方向再从源点求一次即可 spfa代码 #include<iostream> #include<vector> #include<algorithm> #include<cstring> #include<cstdio> #include<cmath> #include<cstdlib>…

POJ 1511 Invitation Cards / UVA 721 Invitation Cards / SPOJ Invitation / UVAlive Invitation Cards / SCU 1132 Invitation Cards / ZOJ 2008 Invitation Cards / HDU 1535 (图论,最短路径) Description In the age of television, not many people attend theater perfor…

Invitation Cards Time Limit: 8000MS   Memory Limit: 262144K Total Submissions: 16178   Accepted: 5262 Description In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They wan…

Invitation Cards 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/J Description In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater a…

In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessa…

Invitation Cards Time Limit: 8000MS   Memory Limit: 262144K Total Submissions: 33435   Accepted: 11104 Description In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They wa…

题意 : 给出 P 个顶点以及 Q 条有向边,求第一个点到其他各点距离之和+其他各点到第一个点的距离之和的最小值 分析 : 不难看出 min( 第一个点到其他各点距离之和+其他各点到第一个点的距离之和 ) = min( 第一个点到其他各点距离之和) + min( 其他各点到第一个点的距离之和 ),前者较为简单,建好图后直接跑一遍最短路即得,关键在于后者怎么方便的求出来.这里利用到了一个逆向思维,如果将所有的有向边反过来,那么就能求出其他点到源点的最短路了,那么只要存储两种边,一个正向边(即题目所…

Invitation Cards Time Limit: 8000MS   Memory Limit: 262144K Total Submissions: 18198   Accepted: 5969 Description In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They wan…

题目传送门 1 2 题意:有向图,所有点先走到x点,在从x点返回,问其中最大的某点最短路程 分析:对图正反都跑一次最短路,开两个数组记录x到其余点的距离,这样就能求出来的最短路以及回去的最短路. POJ 3268 //#include <bits/stdc++.h> #include <cstdio> #include <queue> #include <algorithm> #include <cstring> using namespace…

Invitation Cards Time Limit : 16000/8000ms (Java/Other)   Memory Limit : 524288/262144K (Java/Other) Total Submission(s) : 7   Accepted Submission(s) : 1 Problem Description In the age of television, not many people attend theater performances. Antiq…