起初误以为到每个叶子的概率一样于是.... /* CodeForces 839C - Journey [ DFS,期望 ] | Codeforces Round #428 (Div. 2) */ #include <bits/stdc++.h> using namespace std; const int N = 100005; int n; vector<int> G[N]; double dp[N], val[N]; bool vis[N]; void dfs(int u, i…

There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads. Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But…

C. Journey time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other…

[题目链接] http://codeforces.com/contest/839/problem/C [算法] 概率DP 时间复杂度 : O(N) [代码] #include<bits/stdc++.h> using namespace std; #define MAXN 100010 int tot , n; int head[MAXN]; bool visited[MAXN]; double f[MAXN]; struct edge { int to , nxt; } e[MAXN <…

There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads. Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But…

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Description Little boy Igor wants to become a traveller. At first, he decided to visit all the cities of his motherland - Uzhlyandia. It is widely known that Uzhlyandia has…

$dp$,拓扑排序. 记$dp[i][j]$表示走到节点$i$,走过了$j$个点的最小时间,然后就可以递推了.要注意的是节点$1$的入度一开始不一定等于$0$. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<v…

<题目链接> 题目大意:一个DAG图有n个点,m条边,走过每条边都会花费一定的时间,问你在不超过T时间的条件下,从1到n点最多能够经过几个节点. 解题分析:对这个有向图,我们进行拓扑排序,并且在拓扑排序的过程中,用dp来进行状态的转移,$dp[i][j]$表示,在以$i$为终点的且经过$j$个点的路径中,所花的最少时间. #include <bits/stdc++.h> using namespace std; #define pb push_back ; int dp[N][N]…

题意:给定一个有向图,你从1出发到n,走尽可能多的点,并且使总权值不大于t. 析:在比赛时,竟然看成有向图了,就想了好久,感觉dp,但是不会啊...如果是有向图就好做多了,枚举边,然后打印就好,dp[i][j] 表示, 经过 i 个结点,并且在 j的最小时间. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #inc…

Little boy Igor wants to become a traveller. At first, he decided to visit all the cities of his motherland — Uzhlyandia. It is widely known that Uzhlyandia has n cities connected with m bidirectional roads. Also, there are no two roads in the countr…