题目链接:http://poj.org/problem?id=1511 Invitation Cards Time Limit: 8000MS   Memory Limit: 262144K Total Submissions: 29286   Accepted: 9788 Description In the age of television, not many people attend theater performances. Antique Comedians of Malidine…

题目链接:http://poj.org/problem?id=1511 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 and, most of all, Antique Comedies. They ha…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Invitation Cards Time Limit: 5 Seconds      Memory Limit: 65536 KB In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fa…

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 and, most of all, Antique Comedies. They have printed invitation cards with all…

Invitation Cards Time Limit: 8000MS   Memory Limit: 262144K Total Submissions: 34743   Accepted: 11481 题目链接:http://poj.org/problem?id=1511 Description: In the age of television, not many people attend theater performances. Antique Comedians of Malidi…

题目链接 Problem 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 and, most of all, Antique Comedies. They have printed invitation c…

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…

边取反,从汇点跑单源最短路即可. #include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std; #define MAXN 1111111 #define MAXM 1111111 inline void in(int &ret){ ; '); +c-',c=getchar(); } void out(long long x){ )…

题目链接:https://cn.vjudge.net/problem/POJ-1511 题意 给出一个图 求从节点1到任意节点的往返路程和 思路 没有考虑稀疏图,上手给了一个Dijsktra(按紫书上的存边方法) 直接超时 写了一个极限大小数据 发现读入时间很长,Dij时间也很长,相当于超时超到姥姥家了 赶紧做优化 发现稀疏图,于是换Bellman(spfa) 换邻接表 (虽然没有必要)scanf换成getchar输入模版,大量数据可以节省大概800ms的样子 稀疏图适用Bellman(opti…

<题目链接> 题目大意: 有向图,求从起点1到每个点的最短路然后再回到起点1的最短路之和. 解题分析: 在求每个点到1点的最短路径时,如果仅仅只是遍历每个点,对它们每一个都进行一次最短路算法,那么即使是用了堆优化的dijkstra,时间复杂度也高达$O(n^2log(n))$,而本题有1000000个点,毫无疑问,这种想法必然是不可行的,所以我们可以采用逆向思维,将图中的每一条有向边全部反向,然后以1为起点,仅做一次dijkstra,就能得到1到所有点的最短距离,即反向前的,所有点到1点的最短…