题目链接:http://poj.org/problem?id=3071
题意:
足球赛制是淘汰赛,每次是1和2打比赛,3和4……然后1和2的胜者和3和4的胜者打比赛,i打j的胜率是 Pij,问哪只队伍最有可能夺冠?(即获胜概率最高)。
题解:
简单的概率dp题,只需要判断一下j和k能否碰上即可。
代码:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int inf = 1 << 26;
double f[205][205], a[205][205];
int n;
int main(){
while( scanf("%d", &n), n != -1){
memset(f, 0, sizeof(f));
for ( int i = 0; i < (1<<n); i ++ ) {
for ( int j = 0; j < (1<<n); j ++ ) scanf("%lf", &a[i][j]);
}
for ( int i = 0; i < (1<<n); i ++ ) {
f[0][i] = 1;
}
for ( int i = 1; i <= n; i ++) {
for ( int j = 0; j < (1<<n); j ++ ) {
for ( int k = 0; k< (1<<n); k ++ ) {
if((k>>(i-1)^1) == (j >> (i-1))) {
f[i][j] += a[j][k]*f[i-1][j]*f[i-1][k];
}
}
}
}
int ans = 0;
for ( int i = 0; i < (1<<n); i ++ ) {
if(f[n][i] > f[n][ans]) {
ans = i;
}
}
printf("%d\n", ans+1);
}
return 0;
}