PAT 1079

傻逼题...

PAT的出题我一直很想吐槽，数据之类的不是很灵，该给过不过，不该过随意过。

这道题需要全员double才能过，需要注意的是如果在float后面加上f也是过不了的... 变量都是double接着了还是过不了是为什么，感觉好奇怪，有可能是二进制表达的关系，大数据有凑巧是0.5或者是0.25的所以改成double但是没改0.01后面的f结果就过了的...

简单BFS

#include <vector>
#include <iostream>
#include <queue>
#include <iomanip>

using namespace std;

double bfs(const vector<vector<int> > &graph, const double p, const double r){
    double p_fac = (double)(1.0 + r * 0.01);
    double cur_price = p;
    double res = 0.0f;

    queue<int> cur_level;
    cur_level.push(0);
    while (!cur_level.empty()){
        queue<int> next_level;

        while (!cur_level.empty()){
            int cur_idx = cur_level.front();
            cur_level.pop();

            int size = graph[cur_idx][0];
            if (size == 0){
                res += graph[cur_idx][1] * cur_price;
            } else {
                for (int i = 1; i < size + 1; i++){
                    next_level.push(graph[cur_idx][i]);
                }
            }
        }

        cur_price *= p_fac;
        cur_level = next_level;
    }

    return res;
}

int main(){
    int N;
    double P, r;

    cin >> N >> P >> r;

    vector<vector<int> > graph(N, vector<int>());
    for (int i = 0; i < N; i++){
        int cnt, num;
        cin >> cnt;
        graph[i].push_back(cnt);

        if (cnt == 0){
            cin >> num;
            graph[i].push_back(num);
        } else {
            for (int j = 0; j < cnt; j++){
                cin >> num;
                graph[i].push_back(num);
            }
        }
    }

    double res = bfs(graph, P, r);

    cout << fixed << setprecision(1) << res << endl;

    return 0;
}