hdu - 随笔分类 - see_why

2112HDU Today(Dijkstra)

摘要：#include <iostream>#include <algorithm>#include <cstring>#include <vector>#include <string>#include <map>#include <queue>using namespace std;#ifndef ONLINE_JUDGE#include <fstream>ifstream fin("test.txt");#define cin fin#endifconst int INF = 1 阅读全文

posted @ 2013-05-24 20:52 see_why 阅读(199) 评论(0) 推荐(0)

2112HDU Today(SPFA)

摘要：刚开始用spfa(久闻大名却一次都没用过）过不了，改dijkstra，wa了十几次后，心灰意冷。。1个小时后终于发现是判断起点与终点相同时，直接continue了，导致后面的数据没有完全输入...改正这个问题后spfa的代码也过了，哈哈#include <iostream>#include <algorithm>#include <cstring>#include <vector>#include <string>#include <map>#include <queue>using namespace std 阅读全文

posted @ 2013-05-24 20:51 see_why 阅读(163) 评论(0) 推荐(0)

1162Eddy's picture

摘要：一次通过，刚学prim的时候做过一道类似的，将每个点之间的距离保存在double型的邻接矩阵里，然后调用prim#include <iostream>#include <algorithm>#include <iomanip>#include <cstring>#include <cmath>using namespace std;#ifndef ONLINE_JUDGE#include <fstream>ifstream fin("test.txt");#define cin fin#endifdou 阅读全文

posted @ 2013-05-23 20:13 see_why 阅读(161) 评论(0) 推荐(0)

1869六度分离

摘要：实在是太粗心了，floyd算法中把k打成了i，找了20分钟才找到错.... 题目不难，最多用6个人就能联系在一起，等价于任意2点的距离不超过7#include <iostream>#include <algorithm>#include <cstring>using namespace std;#ifndef ONLINE_JUDGE#include <fstream>ifstream fin("test.txt");#define cin fin#endifint graph[110][110],n,m;const int 阅读全文

posted @ 2013-05-23 19:06 see_why 阅读(182) 评论(0) 推荐(0)

1272小希的迷宫

摘要：思路是用并查集判断连通，然后判断边数+1==点数，就是判断是否为一颗生成树#include <iostream>#include <set>#include <algorithm>using namespace std;#ifndef ONLINE_JUDGE#include <fstream>ifstream fin("test.txt");#define cin fin#endifconst int MAXN = 100010;int p[MAXN],vis[MAXN],maxv,edge,ok;set<int> 阅读全文

posted @ 2013-05-23 15:18 see_why 阅读(181) 评论(0) 推荐(0)

hdu1116Play on Words

摘要：第一次接触欧拉路，看了大神的解题报告后写出并查集+欧拉回路，先要用并查集判断是否连通，然后再根据每个节点的入度与出度判断是否为欧拉回路或欧拉路如果所有点入度等于出度，则为欧拉回路如果起点的入度比出度小一且终点的入度比出度大一，则为欧拉路#include <iostream>#include <cstring>using namespace std;#ifndef ONLINE_JUDGE#include <fstream>ifstream fin("test.txt");#define cin fin#endifint p[27],vis 阅读全文

posted @ 2013-05-22 20:48 see_why 阅读(130) 评论(0) 推荐(0)

3549Flow Problem

摘要：[cce_cpp]//网络流，基本是照着书上的代码敲的，还敲错了o(╯□╰)o#include <cstdio>#include <algorithm>#include <queue>#include <cstring>using namespace std;int cap[20][20],flow[20][20],a[20],p[20];const int INF = 1000000;int main(){ int t,n,m,x,y,c,f; scanf("%d",&t); for(int i = 1; i < 阅读全文

posted @ 2013-05-22 14:50 see_why 阅读(175) 评论(0) 推荐(0)

hdu2544

摘要：#include <iostream> //第一道用Dijkstra解的题#include <cstring>using namespace std;#ifndef ONLINE_JUDGE#include <fstream>ifstream fin("test.txt");#define cin fin#endifint n,m;const int INF = 1000010;int vis[200],dis[200];int graph[200][200];int main(){ ios::sync_with_stdio(false) 阅读全文

posted @ 2013-05-19 13:16 see_why 阅读(150) 评论(0) 推荐(0)

hdu1874

摘要：[cce_cpp]#include <iostream>#include <algorithm>using namespace std;#ifndef ONLINE_JUDGE//最开始想用来练习spfa+邻接表，老是wa,改用floyd + 邻接矩阵，还是wa，最后无奈看了discuss,//才发现忘记判断起点与终点相同的情况...#include <fstream>ifstream fin("test.txt");#define cin fin#endifconst int INF = 99999999;int graph[220][ 阅读全文

posted @ 2013-05-19 12:28 see_why 阅读(251) 评论(0) 推荐(0)

hdu1232 畅通工程

摘要：#include <iostream> #include <fstream>using namespace std;#ifndef ONLINE_JUDGEifstream fin("test.txt");#define cin fin#endifint bin[1010];int findx(int x){ int r = x; while(bin[r] != r) { r = bin[r]; } return r;}void merge(int x,int y){ bin[x] = y;}int main(){ int n,m,a,b... 阅读全文

posted @ 2013-02-04 21:22 see_why 阅读(183) 评论(0) 推荐(0)

hdu1097A hard puzzle

摘要：#include <iostream>using namespace std;int main(){ long long n,m,i,mul; while(cin >> n >> m) { mul = 1; m = m % 4; //找规律，一个数的n次方。。它的个位数四次一循环 if(!m) m += 4; for(i = 0 ; i < m ; ++i) { mul = mul * n % 10; } cout << mul ... 阅读全文

posted @ 2013-01-28 20:36 see_why 阅读(145) 评论(0) 推荐(0)

hdu1871无题

摘要：#include <iostream>#include <algorithm>using namespace std;struct hotel{ int num; int room_num; int price;};bool cmp(hotel a,hotel b) //先按价格升序，再按房间数升序，最后按编号{ if(a.price != b.price) return a.price < b.price; else if(a.room_num != b.room_num) return a.price < b.price; else ... 阅读全文

posted @ 2013-01-28 20:32 see_why 阅读(131) 评论(0) 推荐(0)

hdu4492 Mystery

摘要：//杭电的一道月赛题，当时也做过了//第一次提交用c++写，虽然过了，但感觉效率好低，然后用c重写了一遍。。。#include <stdio.h>#include <string.h> int main(){ char a[100]; int p,d,n,x,sum,m; scanf("%d",&p); while(p--) { sum = 0; scanf("%d",&d); getchar(); gets(a); scanf("%d",&n); printf("%d &qu 阅读全文

posted @ 2013-01-25 22:22 see_why 阅读(247) 评论(0) 推荐(0)

1027Ignatius and the Princess II

摘要：Ignatius and the Princess IITime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3027Accepted Submission(s): 1818Problem DescriptionNow our hero finds the door to the BEelzebub feng5166. He opens the door and finds feng5166 is about to kill our pretty P 阅读全文

posted @ 2013-01-24 17:00 see_why 阅读(180) 评论(0) 推荐(0)

1236排名

摘要：Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11054Accepted Submission(s): 4168Problem Description今天的上机考试虽然有实时的Ranklist，但上面的排名只是根据完成的题数排序，没有考虑每题的分值，所以并不是最后的排名。给定录取分数线，请你写程序找出最后通过分数线的考生，并将他们的成绩按降序打印。Input测试输入包含若干场考试的信息。每场考试信息的第1行给出考生人数N ( 0 < N 阅读全文

posted @ 2013-01-21 20:50 see_why 阅读(187) 评论(0) 推荐(0)

hdu_acm2010水仙花数

摘要：水仙花数Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 62100Accepted Submission(s): 18246Problem Description春天是鲜花的季节，水仙花就是其中最迷人的代表，数学上有个水仙花数，他是这样定义的：“水仙花数”是指一个三位数，它的各位数字的立方和等于其本身，比如：153=1^3+5^3+3^3。现在要求输出所有在m和n范围内的水仙花数。Input输入数据有多组，每组占一行，包括两个整数m和n（100 阅读全文

posted @ 2012-11-24 21:27 see_why 阅读(274) 评论(0) 推荐(0)

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