HDU4418 Time travel(期望dp 高斯消元)

题意

题目链接

Sol

mdzz这题真的太恶心了。。

首先不难看出这就是个高斯消元解方程的板子题

(f[x] = sum_{i = 1}^n f[to(x i)] * p[i] ave)

(ave)表示每次走的期望路程

然后一件很恶心的事情是可以来回走，而且会出现(M > N)的情况(因为这个调了两个小时。。)

一种简单的解决方法是在原序列的后面接一段翻转后的序列

比如(1 2 3 4)可以写成(1 2 3 4 3 2)

然后列式子解方程就行了

附送一个数据生成器

代码语言：javascript复制

#include<bits/stdc  .h>
using namespace std;
int main() {
    freopen("a.in", "w", stdout);
    srand((unsigned)time(NULL));
    int T = 30;
    printf("%dn", T);
    while(T--) {
        int N = rand() % 100   1, M = rand() % 20   1, Y = rand() % N, X = rand() % N, D = rand() % 2;
        if(X == 0 || X == N - 1) D = -1;
        printf("%d %d %d %d %dn", N, M, Y, X, D);
        int res = 100;
        for(int i = 1; i <= M - 1; i  ) {
            int rd;
            if(res == 0) rd = 0;
            else rd = rand() % res   1;
            printf("%d ", rd); res -= rd;
        }
        printf("%dn", res);
    }
    return 0;
}

代码语言：javascript复制

#include<bits/stdc  .h> 
#define LL long long 
using namespace std;
const int MAXN = 1001, mod = 998244353;
const double eps = 1e-9;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10   c - '0', c = getchar();
    return x * f;
}
int N, M, X, Y, D, Lim, vis[MAXN];
double g[MAXN][MAXN], p[MAXN], ave;
int Gauss() {
    for(int i = 1; i < Lim; i  ) {
        int mx = i;
        for(int j = i   1; j < Lim; j  ) 
            if(fabs(g[j][i]) > fabs(g[mx][i])) mx = j;
        swap(g[i], g[mx]);
        //assert(g[i][i] > eps);
        if(fabs(g[i][i]) < eps) return -1;
        for(int j = i   1; j < Lim; j  ) {
            double p = g[j][i] / g[i][i];
            for(int k = i   1; k <= Lim; k  ) 
                g[j][k] -= g[i][k] * p;
        }
    }   
    for(int i = 1; i < Lim; i  ) if(fabs(g[i][i]) < eps) return -1;
    for(int i = Lim - 1; i >= 1; i--) {
        g[i][i] = g[i][Lim] / g[i][i];
        for(int j = i - 1; j >= 1; j--)
            g[j][Lim] -= g[j][i] * g[i][i], g[j][i] = 0;
    }
}
int walk(int a, int b) { 
    b %= (Lim - 1);
    int x = a   b;
    if(x <= Lim - 1) return x;
    return x % (Lim - 1);
}
void init() {
    memset(g, 0, sizeof(g));
    memset(vis, 0, sizeof(vis));
    ave = 0;
}
void BFS() {
    queue<int> q; q.push(X); vis[X] = 1;
    while(!q.empty()) {
        int x = q.front(); q.pop();
        for(int i = 1; i <= M; i  ) {
            if(p[i] > eps) {
                int t = walk(x, i);
                if(!vis[t]) q.push(t), vis[t] = 1;
            }
        }
    }
}
void solve() {
    init();
    N = read(); M = read(); Y = read()   1; X = read()   1; D = read();
    Lim = (N << 1) - 1;
    for(int i = 1; i <= M; i  ) p[i] = (double) read() / 100, ave  = (double) i * p[i];
    if(X == Y) {puts("0.00"); return;}
    if(D > 0 || (D == -1 && X > Y)) X = N - X   1, Y = N - Y   1;
    BFS();
    for(int i = 1; i <= 2 * N - 2; i  ) {
        g[i][i] = 1; 
        if(!vis[i]) {g[i][Lim] = 3e18; continue;}
        if(i == Y || (Lim - i   1 == Y)) continue;
        g[i][Lim] = ave;
        for(int j = 1, t; j <= M; j  ) {
            t = walk(i, j);
            g[i][t] -= p[j];
        }
    }
    if((!vis[Y] && !vis[Lim - Y   1]) || (Gauss() == -1)) puts("Impossible !");
    else printf("%.2lfn", g[X][X]);
}
int main() {
    //freopen("a.in", "r", stdin);
    //freopen("b.out", "w", stdout);
    for(int T = read(); T; T--, solve());
    return 0;
}

sum

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