CCF 201503-5 最小花费 传送门
毕竟是个蒟蒻, 所以这CCF-CSP第五题也只能用用O(nm)的算法暴力去解了, 而且常数还TM很大. 得了30分, 也算是出题人放了个小水.
思路就是:
用DFS找到两端点的路径, 同时存储边的长度. O(n)
对每一条路径点判断物价大小,维持当前最小的物价,购买尽可能多的粮食直到可以走到更小的物价处
注意用long long
, 不然你只能得个10分咧
看大佬的满分代码. 点俺点俺
好歹是第五题, 贴出30
分代码也不丢人是不
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
struct Node {
long long from, to, cost;
Node (long long f, long long t, long long c) : from(f), to(t), cost(c) {}
};
const long long maxn = 100000 + 5;
long long n, m, cost[maxn];
vector<Node> G[maxn];
vector<long long> Path, path, Dis, dis; // 真假path
bool vis[maxn];
void DFS(long long from, long long to)
{
if (from == to) {
Path = path;
Dis = dis;
return;
}
for (long long i = 0, v, d; i < G[from].size(); ++i) {
v = G[from][i].to;
d = G[from][i].cost;
if (vis[v]) continue;
vis[v] = true;
path.push_back(v);
dis.push_back(d);
DFS(v, to);
path.pop_back();
dis.pop_back();
}
}
int main()
{
ios::sync_with_stdio(false);
cin >> n >> m;
for (long long i = 1; i <= n; ++i)
cin >> cost[i];
for (long long i = 1, u, v, c; i < n; ++i) {
cin >> u >> v >> c;
G[u].push_back(Node(u, v, c));
G[v].push_back(Node(v, u, c));
}
for (long long i = 0, from, to; i < m; ++i) {
cin >> from >> to;
memset(vis, 0, sizeof(vis));
path.clear();
dis.clear();
path.push_back(from);
vis[from] = true;
DFS(from, to);
long long _min = cost[from], cos = 0, ans = 0;
for (long long i = 0; i < Path.size() - 1; ++i) {
if (cost[Path[i]] < _min) {
ans += cos*_min;
cos = 0;
_min = cost[Path[i]];
}
cos += Dis[i];
}
ans += _min*cos;
cout << ans << endl;
}
}