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POJ 3071 Football (概率DP)
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; }
POJ 3071 Football (概率DP)
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