Codeforces Beta Round #16 (Div. 2 Only) E. Fish 题目链接:## 点击打开链接 题意: 有 \(n\) 条鱼,每两条鱼相遇都会有其中一只吃掉对方,现在给你一个 \(n * n\)的矩阵,表示 \(i\) 吃掉 \(j\) 的概率,最后问你每条鱼存活的概率. 题解: 最多有 \(18\) 条鱼,吃掉的概率都不一样,可以用状态压缩,设\(dp[1<<n]\)种状态,最多有 \(1<<18\) 种状态. $ dp[i]$ 表示当前鱼的状态为…

Codeforces Beta Round #16 (Div. 2 Only) http://codeforces.com/contest/16 A 水题 #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define maxn 500005 typedef long lo…

C. Longest Regular Bracket Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/5/C Description This is yet another problem dealing with regular bracket sequences. We should remind you that a bracket sequence…

题意: 有n种卡片,每包面里面,可能有一张卡片或没有,已知每种卡片在面里出现的概率,求获得n种卡片,需要吃面的包数的期望 分析: n很小,用状压,以前做状压时做过这道题,但概率怎么推的不清楚,现在看来就是基本的概率dp dp[s]表示获得卡片种数情况是s时期望包数,dp[(1<<n)-1]=0,dp[0]就是答案 dp[s]=sum(dp[s+(1<<j)]*p[j])+1+(1-tmp)*dp[s](tmp是未吃到的卡片的概率和) 移项化简即可 #include <map&…

C. Monitor time limit per test 0.5 second memory limit per test 64 megabytes input standard input output standard output Reca company makes monitors, the most popular of their models is AB999 with the screen size a × b centimeters. Because of some pr…

D. Two Paths 题目连接: http://codeforces.com/contest/14/problem/D Description As you know, Bob's brother lives in Flatland. In Flatland there are n cities, connected by n - 1 two-way roads. The cities are numbered from 1 to n. You can get from one city t…

Two Paths time limit per test 2 seconds memory limit per test 64 megabytes input standard input output standard output As you know, Bob's brother lives in Flatland. In Flatland there are n cities, connected by n - 1 two-way roads. The cities are numb…

题意: 有n个节点的图,开始有一些边存在,现在每天任意选择两点连一条边(可能已经连过),求使整个图联通的期望天数. 分析: 由于开始图可以看做几个连通分量,想到了以前做的一个题,一个点代表一个集合(这里是连通分量)进行压缩 dp[i][s]表示最后连接的第i个联通分量,联通状态是s时的期望天数,dp[0][1],即为答案,由于s可能很大,用记忆化搜索 #include <map> #include <set> #include <list> #include <c…

Codeforces Beta Round #83 (Div. 1 Only) A. Dorm Water Supply 题意 给你一个n点m边的图,保证每个点的入度和出度最多为1 如果这个点入度为0,那么这个点就是水龙头点. 如果这个点的出度为0,那么这个点就是储存点. 现在让你把所有水龙头到储存点的路径都输出出来,且输出这条路径的边权最小值 题解 显然是个仙人掌图,所以直接XJB暴力就好了 代码 #include<bits/stdc++.h> using namespace std; co…

Codeforces Beta Round #72 (Div. 2 Only) http://codeforces.com/contest/84 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplac…