POJ 3071 Football (概率DP)

Posted 2020-08-15 dwtfukgv

题意：给定 2的n次方 个团队对每个队的战胜的概率，一块要打 n 场，每场都是 1 对 2， 2 对 3，每次都取赢的一方，问你最后谁是冠军的概率最大。

析：dp[i][j] 表示 第 i 场 j 胜的概率，每次只要算 i 相邻的且不是已经打过的 2 i-1次方个队，最后再选出概率最大的就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e4 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;

}
double a[150][150];
double dp[10][150];
int f[10];

int main(){
    f[0] = 1;
    for(int i = 1; i < 10; ++i) f[i] = f[i-1] * 2;
    for(int i = 0; i < 150; ++i)  dp[0][i] = 1.0;
    while(scanf("%d", &n) == 1 && n >= 0){
        for(int i = 1; i <= f[n]; ++i)
            for(int j = 1; j <= f[n]; ++j)
                scanf("%lf", &a[i][j]);

        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= f[n]; ++j)
                dp[i][j] = 0.0;
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= f[n]; j += f[i])
                for(int k = j; k < j+f[i-1]; ++k)
                    for(int l = j+f[i-1]; l < j+f[i-1]+f[i-1]; ++l){
                        dp[i][k] += dp[i-1][k] * dp[i-1][l] * a[k][l];
                        dp[i][l] += dp[i-1][k] * dp[i-1][l] * a[l][k];
                    }

        double ans = 0.0;
        int idx = -1;
        for(int i = 1; i <= f[n]; ++i) if(dp[n][i]-ans >= eps){
            ans = dp[n][i];
            idx = i;
        }
        printf("%d\n", idx);
    }
    return 0;
}