http://poj.org/problem?id=3071 (题目链接) 题意 ${2^n}$个队伍打淘汰赛,输的被淘汰.第1个队打第2个队,第3个队打第4个队······给出第i个队伍打赢第j个队伍的概率p[i][j],求哪只队伍获得冠军的可能性最大 Solution 很简单,想到一个dp方程:${f_{i,j}}$表示第i轮,j胜出的概率.转移很显然: $${f_{i,j}=f_{i-1,j}×f_{i-1,k}×p_{j,k}}$$ 代码 // poj3071 #include<algo…

poj3071 Football 题意:有2^n支球队比赛,每次和相邻的球队踢,两两淘汰,给定任意两支球队相互踢赢的概率,求最后哪只球队最可能夺冠. 我们可以十分显然(大雾)地列出转移方程(设$f[ i ][ j ]$为第 $j$ 支球队踢赢第 $i$ 场比赛的概率,$k$为枚举的对手): $f[ i ][ j ] += f[ i-1 ][ j ] * f[ i-1 ][ k ] * p[ j ][ k ]$ 现在问题来了:怎么枚举 k, k的范围是啥 我们先列出比赛的模型(a Tower,n=…

Description Consider a single-elimination football tournament involving 2n teams, denoted 1, 2, …, 2n. In each round of the tournament, all teams still in the tournament are placed in a list in order of increasing index. Then, the first team in the l…

Football Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3469   Accepted: 1782 Description Consider a single-elimination football tournament involving 2n teams, denoted 1, 2, …, 2n. In each round of the tournament, all teams still in the…

学习位运算在比赛的技巧 http://poj.org/problem?id=3071 Football Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3075   Accepted: 1558 Description Consider a single-elimination football tournament involving 2n teams, denoted 1, 2, …, 2n. In each roun…

题目大意,1<<n个球队比赛赛程是这样的 1 1    1 1   1 1  1 另dp[i][k]为k队进入第i场的概率 #include<iostream> #include<string.h> #include<stdio.h> using namespace std; <<); double a[maxa][maxa]; ][maxa]; int main(){ int n; while(scanf("%d", &am…

Description Consider a single-elimination football tournament involving 2n teams, denoted 1, 2, …, 2n. In each round of the tournament, all teams still in the tournament are placed in a list in order of increasing index. Then, the first team in the l…

                                                                                             Football Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2590   Accepted: 1315 Description Consider a single-elimination football tournament inv…

http://poj.org/problem?id=3071 题意:有2^n个队伍,给出每两个队伍之间的胜率,进行每轮淘汰数为队伍数/2的淘汰赛(每次比赛都是相邻两个队伍进行),问哪只队伍成为冠军概率最大. 很基础. 代码 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> #include<queue> u…

题意:n支队伍两两进行比赛,求最有可能获得冠军的队伍. 解题关键:概率dp,转移方程:$dp[i][j] +  = dp[i][j]*dp[i][k]*p[j][k]$表示第$i$回合$j$获胜的概率,原理为全概率公式. 如何判断相邻,通过位运算. 概率dp的过程就像是模拟的过程 复杂度:$O(n{2^n}{2^n})$ #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib&…