LightOJ 1065 Island of Survival (概率DP?)
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题意:有 t 只老虎,d只鹿,还有一个人,每天都要有两个生物碰面,
1.老虎和老虎碰面,两只老虎就会同归于尽
2.老虎和人碰面或者和鹿碰面,老虎都会吃掉对方
3.人和鹿碰面,人可以选择杀或者不杀该鹿
4.鹿和鹿碰面,没事
问人存活下来的概率
析:最后存活肯定是老虎没了,首先可以用概率dp来解决,dp[i][j] 表示 还剩下 i 考虑, j 只鹿存活的概率是多少。
然后每次分析这几种情况即可。
还有一种思路就是只要考虑老虎没了,只要老虎没了就能存活,只要计算老虎全死完的概率就好,首先如果老虎是奇数,是肯定死不完的。老虎是偶数才有可能死完。
代码如下:
概率DP:
#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> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; 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 bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } double dp[maxn][maxn]; int main(){ int T; cin >> T; for(int kase = 1; kase <= T; ++kase){ scanf("%d %d", &n, &m); memset(dp, 0, sizeof dp); dp[n][m] = 1.0; for(int i = n; i; --i) for(int j = m; j >= 0; --j){ double sum = i*(i-1)/2 + i*j + i; if(i >= 2) dp[i-2][j] += dp[i][j]*i*(i-1)/2.0/sum; if(i > 0 && j > 0) dp[i][j-1] += dp[i][j]*i*j/sum; } double ans = 0.0; for(int i = 0; i <= m; ++i) ans += dp[0][i]; printf("Case %d: %.10f\n", kase, ans); } return 0; }
#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> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; 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 bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } double solve(int x){ if(x & 1) return 0.0; double ans = 1.0; while(x){ ans *= (x-1.0) / (x+1.0); x -= 2; } return ans; } int main(){ int T; cin >> T; for(int kase = 1; kase <= T; ++kase){ scanf("%d %d", &n, &m); printf("Case %d: %.10f\n", kase, solve(n)); } return 0; }
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