HDU 4889 Scary Path Finding Algorithm】的更多相关文章

其实这个题是抄的题解啦…… 题解给了一个图,按照那个图模拟一遍大概就能理解了. 题意: 有一段程序,给你一个C值(程序中某常量),让你构造一组数据,使程序输出"doge" 那段代码大概是 SPFA的SLF优化.其实题目的意思是让我们构造一组数据,使得总的出队次数大于C.        数据范围 C<=23,333,333.输出的图中最多有100个点,没有重边.自环.负环. 思路: SLF: 设队首元素为 i, 队列中要加入节点 j, 在        时加到队首而不是队尾, 否则…
Fackyyj loves the challenge phase in TwosigmaCrap(TC). One day, he meet a task asking him to find shortest path from vertex 1 to vertex n, in a graph with at most n vertices and m edges. (1 ≤ n ≤ 100,0 ≤ m ≤ n(n-1)) Fackyyj solved this problem at fir…
(THIS BLOG WAS ORIGINALLY WRTITTEN IN CHINESE WITH LINK: http://www.cnblogs.com/waytofall/p/3732920.html) Foreword: Floyd-Warshall is a classical dynamical programming algorithm for deriving shortest paths between each pair of nodes on a graph. It ha…
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1973 Prime Path Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices…
Shortest Path 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5636 Description There is a path graph G=(V,E) with n vertices. Vertices are numbered from 1 to n and there is an edge with unit length between i and i+1 (1≤i<n). To make the graph more in…
huffman coding, greedy algorithm. std::priority_queue, std::partition, when i use the three commented lines, excution time increase to 15ms from 0ms, due to worse locality? thanks to http://acm.hdu.edu.cn/discuss/problem/post/reply.php?action=support…
题目链接:pid=3631" style="font-size:18px">http://acm.hdu.edu.cn/showproblem.php?pid=3631 Shortest Path Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3962    Accepted Submission(s): 9…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1973 Prime Path Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description The ministers of the cabinet were quite upset by the message from the Chief of Secu…
floyd算法好像很奇妙的样子.可以做到每次加入一个点再以这个点为中间点去更新最短路,效率是n*n. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> using namespace std; ; const int INF = 0x7FFFFFFF; int A[maxn][maxn], flag[maxn]; int…
http://acm.hdu.edu.cn/showproblem.php?pid=5385 题意: 给定一张n个点m条有向边的图,构造每条边的边权(边权为正整数),令d(x)表示1到x的最短路,使得存在点i(1<=i<=n)满足d(1)<d(2)<…<d(i)>d(i+1)>…>d(n). 从两边向中间构造. 开始L=1,R=n 从L开始bfs,顺次构造L,L+1,L+2…… 构造不动了再从R开始bfs,顺次构造R,R-1,R-2…… 然后在从L开始………