POJ 3071 Football （概率dp）

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题目链锁：http://poj.org/problem?id=3071

题意：有2^n个队伍，进行n轮比赛，按编号顺序进行比赛，如第一支对第二支，第三支对第四支……，给出队伍i打败队伍j的概率，求胜率最高的那只队。

dp[i][j]表是第i轮第j队胜出，

第n轮获胜的队是由l = 1到r = 2^n 角逐出来的，那么判断j是在m=(l+r)/2左边还是右边，

如果在左边则dp[i][j] = dp[i-1][j] * d[i-1][k] * p[i][k] , m+1 <= k <= r;

如果在右边则dp[i][j] = dp[i-1][j] * d[i-1][k] * p[i][k] , l <= k <= m;

接着把l,r递归下去;

边界是i == 0 返回1

#include<iostream>#include<cstring>#include<string>#include<cstdio>#include<stdio.h>#include<algorithm>#include<cmath>#include<set>#include<map>#include<queue>using namespace std;#define inf 0x3f3f3f3f#define eps 1e-9#define mod 10007#define FOR(i,s,t) for(int i = s; i < t; ++i )#define REP(i,s,t) for( int i = s; i <= t; ++i )#define LL long long#define ULL unsigned long long#define pii pair<int,int>#define MP make_pair#define lson id << 1 , l , m#define rson id << 1 | 1 , m + 1 , r #define maxn ( 1000+10 )#define maxe ( 20000+10 )#define mxn 20000double dp[10][maxn];double p[maxn][maxn];bool vis[11][maxn];double DP ( int i, int j, int l, int r  ) {if( i == 0 ) return 1;if( vis[i][j] ) return dp[i][j];vis[i][j] = 1;int m = l + r >> 1;if( j <= m )REP( k ,m+1, r )dp[i][j] += DP( i-1, j, l , m ) * DP( i-1, k, m+1, r ) * p[j][k];else REP( k, l, m )dp[i][j] += DP( i-1, j, m+1, r ) * DP( i-1, k, l, m ) * p[j][k];return dp[i][j];}int main () {int n;while( ~scanf("%d", &n ) ) {if( n == -1 ) break;memset( vis,0, sizeof( vis ) );memset( dp, 0, sizeof( dp ) ) ;int m = ( 1 << n );for( int i = 1; i <= m; ++i ) for ( int j = 1; j <= m; ++j ) scanf("%lf", &p[i][j] );double mx = 0, ans;int res;for( int i = 1; i <= m; ++i ) {ans = DP( n, i, 1, 1<<n );if( ans > mx )res = i, mx = ans;}//cout<<mx<<endl;printf("%d\n", res );}}

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