hdu 5001 Walk(记忆化搜索)

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题目链接：hdu 5001 Walk

题目大意：给定一张图，每次随机移动到下一个节点，问所走完d步后，每个点没有被走过的概率。

解题思路：枚举每节点x，dp[i][j]表示第i步走到j节点的概率（没有经过x）。

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int maxn = 55;const int maxm = 10005;int N, M, D;vector<int> g[maxn];double dp[maxm][maxn];void init () {    scanf("%d%d%d", &N, &M, &D);    for (int i = 0; i <= N; i++)        g[i].clear();    int x, y;    for (int i = 0; i < M; i++) {        scanf("%d%d", &x, &y);        g[x].push_back(y);        g[y].push_back(x);    }    for (int i = 1; i <= N; i++)        g[0].push_back(i);}double solve (int u) {    double ret = 0;    memset(dp, 0, sizeof(dp));    dp[0][0] = 1;    for (int i = 0; i <= D; i++) {        for (int j = 0; j <= N; j++) {            if (u == j)                continue;            double p = 1.0 / g[j].size();            for (int k = 0; k < g[j].size(); k++)                dp[i+1][g[j][k]] += dp[i][j] * p;        }        ret += dp[i+1][u];    }    return 1.0 - ret;}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        init();        for (int i = 1; i <= N; i++)            printf("%.10lf\n", solve(i));    }    return 0;}

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