HDU 3853 LOOPS (期望DP)

Posted 2020-08-13 dwtfukgv

题意：给定一个 n * m的矩阵，然后你从 （1，1）到 （n，m），每次你有三种可能，不动，向右，向下，每次要消耗2个魔法，并且给定每个概率，

问你走出去的期望。

析：dp[i][j] 表示从 （i，j）到终点的概率。然后一路逆推回去就好。

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>
//#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 = 1e3 + 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 dp[maxn][maxn];
double a[maxn][maxn][3];

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < m; ++j)
                scanf("%lf %lf %lf", &a[i][j][0], &a[i][j][1], &a[i][j][2]);
        dp[n-1][m-1] = 0.0;
        for(int i = n-1; i >= 0; --i)
            for(int j = m-1; j >= 0; --j){
                if(i == n-1 && j == m-1)  continue;
                if(fabs(1.0 - a[i][j][0]) < eps)  continue;
                dp[i][j] = ((2+dp[i][j+1])*a[i][j][1] + (2+dp[i+1][j])*a[i][j][2] + 2*a[i][j][0]) / (1.0-a[i][j][0]);
            }

        printf("%.3f\n", dp[0][0]);
    }
    return 0;
}