HDU5001-Walk(记忆化搜索)

来源：互联网发布：万花捏脸数据编辑：程序博客网时间：2024/05/17 09:14

题目链接

题意：一张无向图，要你求出走d步之后，每个点无法到达的概率。

思路：记忆化搜索，枚举每个点a，dp[i][j]表示走了i步到达j点的概率(不包括a点)，注意初始化清空。

代码：

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

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