PAT天梯赛 L2-020. 功夫传人 【DFS】

题目链接

https://www.patest.cn/contests/gplt/L2-020

思路

从师父开始 一层一层往下搜

然后 搜到 得道者 就更新答案

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = 3.14159265358979323846264338327;
const double E = exp(1);
const double eps = 1e-6;

const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int MOD = 1e9 + 7;

ll n;

double z, r;

struct Node
{
    int tot;
    vector <int> v;
}temp;

ld ans = 0.0;

vector <Node> d, da;

void dfs(int x, int cur)
{
    if (d[x].tot == 0)
    {
        ans += z * pow(r, cur) * d[x].v[0];
        return;
    }
    vector <int>::iterator it;
    for (it = d[x].v.begin(); it != d[x].v.end(); it++)
    {
        dfs((*it), cur + 1);
    }
}

int main()
{
    int num;
    scanf("%lld%lf%lf", &n, &z, &r);
    r = (100 - r) * 1.0 / 100;
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &temp.tot);
        temp.v.clear();
        if (temp.tot == 0)
        {
            scanf("%d", &num);
            temp.v.pb(num);
        }
        for (int j = 0; j < temp.tot; j++)
        {
            scanf("%d", &num);
            temp.v.pb(num);
        }
        d.pb(temp);
    }
    dfs(0, 0);
    printf("%.0Lf\n", ans - 0.5);
}