POJ 3071 Football 概率dp

题目链接：

Football

Time Limit: 1000MS
Memory Limit: 65536K #### 问题描述 > 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 list plays the second team, the third team plays the fourth team, etc. The winners of these matches advance to the next round, and the losers are eliminated. After n rounds, only one team remains undefeated; this team is declared the winner. > > Given a matrix P = [pij] such that pij is the probability that team i will beat team j in a match determine which team is most likely to win the tournament.

输入

The input test file will contain multiple test cases. Each test case will begin with a single line containing n (1 ≤ n ≤ 7). The next 2n lines each contain 2n values; here, the jth value on the ith line represents pij. The matrix P will satisfy the constraints that pij = 1.0 − pji for all i ≠ j, and pii = 0.0 for all i. The end-of-file is denoted by a single line containing the number −1. Note that each of the matrix entries in this problem is given as a floating-point value. To avoid precision problems, make sure that you use either the double data type instead of float.

输出

The output file should contain a single line for each test case indicating the number of the team most likely to win. To prevent floating-point precision issues, it is guaranteed that the difference in win probability for the top two teams will be at least 0.01.

样例输入

2
0.0 0.1 0.2 0.3
0.9 0.0 0.4 0.5
0.8 0.6 0.0 0.6
0.7 0.5 0.4 0.0
-1

样例输出

2

题意

有2^n支队伍打比赛，每次编号靠近的两个人（1,2；3,4；...）要打场比赛，也就是每轮要淘汰掉一半的选手，直到剩下一支队伍。
现在你已经知道对于任意的两支队伍i,j，i打败j的概率是p，j打败i的概率是1-p。问最后获胜概率最大的那支队伍的编号。

题解

这里以比赛的轮次作为阶段很自然，所以我们可以想到状态：dp[i][j]表示第i轮j胜出的概率，则有状态转移：dp[i][j]=sigma(dp[i-1][j]*dp[i-1][k]*pro[j][k])，其中k表示第i轮有机会和j交手的队伍，pro[j][k]表示j打赢k的概率。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-9;

const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=8;

double dp[maxn][1<<maxn];
double G[1<<maxn][1<<maxn];

int n;

int main() {
    while(scf("%d",&n)==1&&n!=-1){
        rep(i,0,(1<<n)){
            rep(j,0,(1<<n)){
                scf("%lf",&G[i][j]);
            }
        }

        clr(dp,0);
        rep(i,0,1<<n) dp[0][i]=1.0;
        for(int i=1;i<=n;i++){
            int mask=(1<<n)-(1<<i);
            rep(j,0,(1<<n)){
                rep(k,0,(1<<n)){
                    if(k==j) continue;
                    //这个两个条件筛出第i轮有机会和j打的人
                    if((j&mask)!=(k&mask)) continue;
                    if(((j&(1<<(i-1)))^(k&(1<<(i-1))))==0) continue;
                    dp[i][j]+=dp[i-1][j]*dp[i-1][k]*G[j][k];
                }
            }
        }

        double ma=0.0;
        int pos=-1;
        rep(i,0,(1<<n)){
            if(ma<dp[n][i]){
                ma=dp[n][i];
                pos=i;
            }
        }

        prf("%d
",pos+1);
    }
    return 0;
}

//end-----------------------------------------------------------------------