https://blog.youkuaiyun.com/lxf90zh-cnhttps://blog.youkuaiyun.com/https://blog.youkuaiyun.com/lxf90/article/details/6443379https://blog.youkuaiyun.com/lxf90/article/details/6443379lxf90Tue, 24 May 2011 21:03:00 +0800https://blog.youkuaiyun.com/lxf90/article/details/6440883https://blog.youkuaiyun.com/lxf90/article/details/6440883lxf90Mon, 23 May 2011 21:53:00 +0800https://blog.youkuaiyun.com/lxf90/article/details/6438168https://blog.youkuaiyun.com/lxf90/article/details/6438168lxf90Sun, 22 May 2011 19:12:00 +0800https://blog.youkuaiyun.com/lxf90/article/details/6437277https://blog.youkuaiyun.com/lxf90/article/details/6437277lxf90Sun, 22 May 2011 00:11:00 +0800https://blog.youkuaiyun.com/lxf90/article/details/6389647https://blog.youkuaiyun.com/lxf90/article/details/6389647lxf90Wed, 04 May 2011 02:20:00 +0800https://blog.youkuaiyun.com/lxf90/article/details/6389616https://blog.youkuaiyun.com/lxf90/article/details/6389616lxf90Wed, 04 May 2011 02:17:00 +0800基本的字典树
判断一组字符串中是否出现自己的前缀子串
#include #include const int MAXN = 100010; struct dicTree { int next[10]; bool isWord; void init(){ memset(next,-1,sizeof(next)); isWord = false; } }]]>https://blog.youkuaiyun.com/lxf90/article/details/6388880https://blog.youkuaiyun.com/lxf90/article/details/6388880lxf90Tue, 03 May 2011 23:37:00 +0800 n >> m; while(m--){ cin >> a >> b; map[a][b] = 1; map[b][a] = -1; } /*for(int i =1; ]]>https://blog.youkuaiyun.com/lxf90/article/details/6388644https://blog.youkuaiyun.com/lxf90/article/details/6388644lxf90Tue, 03 May 2011 21:47:00 +0800求割点
关于被割成几个子网络 还不是很懂， 直接累加？
为什么呢？
留下问题 一定要搞懂
#include #include const int MAXN = 1001; int low[MAXN], dfn[MAXN], ans[MAXN], map[MAXN][MAXN]; int n, cnt; void dfs(int now){ dfn[now] = low[now] = cnt++; fo]]>https://blog.youkuaiyun.com/lxf90/article/details/6388610https://blog.youkuaiyun.com/lxf90/article/details/6388610lxf90Tue, 03 May 2011 21:33:00 +0800边的权为不被抓住的概率，注意是概率，需要求的是能逃跑的最大可能
这题与一般的最短路不同，体现在权值上
所以把一般权值的相加改成相乘，利用floyed的算法框架求解
另外注意精度
#include #include #define MAXN 101 double map[MAXN][MAXN]; //边权用浮点数存 int n, m; void floyed(){ for(int k = 0; k < n;]]>https://blog.youkuaiyun.com/lxf90/article/details/6388592https://blog.youkuaiyun.com/lxf90/article/details/6388592lxf90Tue, 03 May 2011 21:24:00 +0800裸的最短路
spfa 水过
#include #include #include #define INF 0x3f3f3f3f #define MAXN 10001 using namespace std; struct edge{ int u, v, w, next; }e[MAXN]; int first[MAXN], dist[MAXN], ecnt; bool inq[MAXN]; int t]]>https://blog.youkuaiyun.com/lxf90/article/details/6388525https://blog.youkuaiyun.com/lxf90/article/details/6388525lxf90Tue, 03 May 2011 21:16:00 +0800https://blog.youkuaiyun.com/lxf90/article/details/6388506https://blog.youkuaiyun.com/lxf90/article/details/6388506lxf90Tue, 03 May 2011 21:07:00 +0800 
给你n个人的联系情况，对任意一个人，求出这个人发消息到其他n-1个人的时间，得到n-1个时间中的最大值，n个最大值中的最小值就是所求。如果网络不通，那就输出disjoint
Floyed算出任意两个人的最小时间
#include #include using namespace std; const int MAXN = 101; const int INF = 0x7FFF; int m , n; int ]]>https://blog.youkuaiyun.com/lxf90/article/details/6388372https://blog.youkuaiyun.com/lxf90/article/details/6388372lxf90Tue, 03 May 2011 20:54:00 +0800给一个网络， 需要求出割点和被这个割点割成的子网络的个数
#include #include const int MAXN = 1001; int low[MAXN], dfn[MAXN], ans[MAXN], map[MAXN][MAXN]; int n, cnt; void dfs(int now){ dfn[now] = low[now] = cnt++; for(int i = 1; i <= n; ++i){ i]]>https://blog.youkuaiyun.com/lxf90/article/details/6388102https://blog.youkuaiyun.com/lxf90/article/details/6388102lxf90Tue, 03 May 2011 20:29:00 +0800https://blog.youkuaiyun.com/lxf90/article/details/6381440https://blog.youkuaiyun.com/lxf90/article/details/6381440lxf90Sun, 01 May 2011 09:17:00 +0800https://blog.youkuaiyun.com/lxf90/article/details/6381432https://blog.youkuaiyun.com/lxf90/article/details/6381432lxf90Sun, 01 May 2011 09:05:00 +0800并查集+set判断个数
#include #include using namespace std; struct G{ int left; int right; int up; int down; }; const int MAXN = 50; int father[MAXN * MAXN]; //map[i][j] = father[i * n + j]对格子的编号 ]]>https://blog.youkuaiyun.com/lxf90/article/details/6381428https://blog.youkuaiyun.com/lxf90/article/details/6381428lxf90Sun, 01 May 2011 08:58:00 +0800https://blog.youkuaiyun.com/lxf90/article/details/6381422https://blog.youkuaiyun.com/lxf90/article/details/6381422lxf90Sun, 01 May 2011 08:48:00 +0800https://blog.youkuaiyun.com/lxf90/article/details/6378398https://blog.youkuaiyun.com/lxf90/article/details/6378398lxf90Sun, 01 May 2011 01:30:00 +0800https://blog.youkuaiyun.com/lxf90/article/details/6373644https://blog.youkuaiyun.com/lxf90/article/details/6373644lxf90Sat, 30 Apr 2011 06:40:00 +0800一、Bellman-Ford算法
最优性原理
 
它是最优性原理的直接应用，算法基于以下事实：
l          如果最短路存在，则每个顶点最多经过一次，因此不超过n-1条边；
l          长度为k的路由长度为k-1的路加一条边得到；
l          由最优性原理，只需依次考虑长度为1，2，…，k-1的最短路。
适用条件&范围
l          单源最短路径(从源点s到其它所有顶点v);