http://poj.org/problem?

id=3071

大致題意：有2^n個(gè)足球隊(duì)分成n組打比賽。給出一個(gè)矩陣a[][],a[i][j]表示i隊(duì)贏得j隊(duì)的概率。n次比賽的流程像這樣France \'98。

問(wèn)最后哪個(gè)隊(duì)最可能得冠軍。

思路：概率dp問(wèn)題。ans[i][j]表示第i輪中j隊(duì)獲勝的概率。

#include <stdio.h> #include <iostream> #include <algorithm> #include <set> #include <map> #include <vector> #include <math.h> #include <string.h> #include <queue> #include <string> #include <stdlib.h> #define LL long long #define _LL __int64 #define eps 1e-8using namespace std; const int INF = 0x3f3f3f3f; const int maxn = 10;int n; int m; double a[130][130]; double ans[maxn][130]; int pow2[8] = {1,2,4,8,16,32,64,128};void solve() {memset(ans,0,sizeof(ans));int t,r;for(int i = 1; i <= m; i++){if(i&1)ans[1][i] = a[i][i+1];elseans[1][i] = a[i][i-1];}for(int i = 2; i <= n; i++){for(int j = 1; j <= m; j++){r = pow2[i]; //周期if(j % r == 0) t = j/r;else t = j/r+1;if(j%r >= 1 && j%r <= r/2){for(int k = t*r; ; k--){if(k%r != 0 && k%r <= r/2)break;ans[i][j] += ans[i-1][j]*a[j][k]*ans[i-1][k];}}else{for(int k = (t-1)*r+1; ; k++){if(k%r >= r/2+1)break;ans[i][j] += ans[i-1][j]*a[j][k]*ans[i-1][k];}}}}}int main() {while(~scanf("%d",&n)){if(n == -1) break;m = pow2[n];for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)scanf("%lf",&a[i][j]);solve();double res = ans[n][1];int pos = 1;for(int i = 2; i <= m; i++){if(res < ans[n][i]){res = ans[n][i];pos = i;}}printf("%d\n",pos);}return 0; }

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