LightOJ - 1151 Snakes and Ladders

LightOJ - 1151

思路：

将期望dp[x]看成自变量，那么递推式就可以看成方程组，用高斯消元求方程组的解就能求解出期望值

高斯消元求解的过程也是期望逆推的过程，注意边界情况的常数项，是6/d，不是1

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb emplace_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 105;
double A[N][N];
void Gauss(int n) {
    for(int i = 0; i < n; i ++) {
        int r = i;
        for(int j = i + 1; j < n; j ++)
            if(fabs(A[j][i]) > fabs(A[r][i])) r = j;
        if(r != i) for(int j = 0; j <= n; j ++) swap(A[r][j], A[i][j]);

        for(int j = n; j >= i; j --) {
            for(int k = i + 1; k < n; k ++)
                A[k][j] -= A[k][i] / A[i][i] * A[i][j];
        }
    }

    for(int i = n - 1; i >= 0; i --) {
        for(int j = i + 1; j < n; j ++)
            A[i][n] -= A[j][n] * A[i][j];
        A[i][n] /= A[i][i];
    }
}
int T, n, a, b, to[105];
int main() {
    scanf("%d", &T);
    for(int cs = 1; cs <= T; ++cs) {
        scanf("%d", &n);
        for (int i = 1; i <= 100; ++i) to[i] = 0;
        for (int i = 1; i <= n; ++i) scanf("%d %d", &a, &b), to[a] = b;

        for (int i = 0; i <= 100; ++i) for (int j = 0; j <= 100; ++j) A[i][j] = 0;
        for (int i = 1; i <= 100; ++i) {
            A[i-1][i-1] = 1;
            if(to[i]) {
                A[i-1][to[i]-1] = -1;
            }
            else {
                int x = min(6, 100-i);
                for (int j = 1; j <= x; ++j) {
                    A[i-1][i+j-1] = -1.0/x;
                }
                if(i < 100) A[i-1][100] = 6.0/x;
            }
        }
        Gauss(100);
        printf("Case %d: %.10f\n", cs, A[0][100]);
    }
    return 0;
}