题意:给出一个供应链,一个根,·每经过一个节点售价增加百分之r,求最少要花多少钱从零售商得到商品,并求出有多少个零售商提供最低的价格。
思路:bfs。
代码
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
#include <queue>
using namespace std;
const int MAX_N = 100010;
const double EPS = 1e-8;
#define isEqual(x, y) (fabs(x-y)<EPS)
int N, x;
double P, R;
int k;
vector<int> G[MAX_N];
vector<int> ret;
map<int, double> pri;
int inq[MAX_N];
double ans;
int cnt;
void dfs(int r) {
queue<int> que;
memset(inq, 0, sizeof(inq));
inq[r] = 1;
que.push(r);
pri[r] = P;
while(!que.empty()) {
int t = que.front(); que.pop();
for (int i = 0; i < G[t].size(); i++) {
int tmp = G[t][i];
if (!pri.count(tmp)) {
pri[tmp] = pri[t] * (1.0+R);
if (!inq[tmp]) que.push(tmp);
} else {
pri[tmp] = min(pri[tmp], pri[t] * (1.0+R));
if (!inq[tmp]) que.push(tmp);
}
}
inq[t] = 0;
}
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
scanf("%d %lf %lf", &N, &P, &R);
R = 0.01 * R;
for (int i = 0; i < N; i++) {
scanf("%d", &k);
for (int j = 0; j < k; j++) {
scanf("%d", &x);
G[i].push_back(x);
}
if (k == 0) {
ret.push_back(i);
}
}
dfs(0);
ans = -1.0; cnt = 0;
for (int i = 0; i < ret.size(); i++) {
if (isEqual(ans, -1.0)) {
ans = pri[ret[i]]; cnt = 1;
} else {
if (isEqual(ans, pri[ret[i]])) cnt++;
else {
if (ans > pri[ret[i]]) {
cnt = 1;
ans = pri[ret[i]];
}
}
}
}
printf("%.4lf %d\n", ans, cnt);
return 0;
}