题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3665 题意分析:以0为起点,求到Sea的最短路径. 所以可以N为超级汇点,使用floyd求0到N的最短路径. /*Seaside Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1151 Accepted Submission(s): 839 P…

题意:给定一个图,你家在0,让你找出到沿海的最短路径. 析:由于这个题最多才10个点,那么就可以用Floyd算法,然后再搜一下哪一个是最短的. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <io…

Problem Description XiaoY is living in a big city, there are N towns in it and some towns near the sea. All these towns are numbered from 0 to N-1 and XiaoY lives in the town numbered '0'. There are some directed roads connecting them. It is guarante…

Seaside Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1356    Accepted Submission(s): 973 Problem Description XiaoY is living in a big city, there are N towns in it and some towns near the sea…

链接:最短路 A.HDU 2544    最短路 算是最基础的题目了吧.............我采用的是Dijkstra算法....... 代码: #include <iostream> #include <cstring> #include <cstdio> using namespace std; #define inf 0x3f3f3f3f ][],d[],vis[],n,m; int Dijkstra() { memset(vis,,sizeof(vis));…

题目链接:http://icpc.njust.edu.cn/Problem/Hdu/3665/ Floyd是经典的dp算法,将迭代过程分成n个阶段,经过n个阶段的迭代所有点对之间的最短路径都可以求出,时间复杂度是O(n^3). 代码如下: #include<bits/stdc++.h> using namespace std; typedef unsigned int ui; typedef long long ll; typedef unsigned long long ull; #defi…

主题链接: HDU:pid=4430" target="_blank">http://acm.hdu.edu.cn/showproblem.php?pid=4430 ZJU:problemId=4888" target="_blank">http://acm.zju.edu.cn/onlinejudge/showProblem.do? problemId=4888 Problem Description Today is Yukari…

Transportation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3045    Accepted Submission(s): 1318 Problem Description There are N cities, and M directed roads connecting them. Now you want to…

King's Game 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5643 Description In order to remember history, King plans to play losephus problem in the parade gap.He calls n(1≤n≤5000) soldiers, counterclockwise in a circle, in label 1,2,3...n. The firs…

Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194    Accepted Submission(s): 3345 Problem Description 话说上回讲到海东集团面临内外交困,公司的元老也只剩下XHD夫妇二人了.显然,作为多年拼搏的商人,XHD不会坐以待毙的.  一天,当他正在苦思冥想解困良策的…