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L3-028 森森旅游(最短路 + multiset)

作者:互联网

题面链接

思路:
由题意得每次兑换旅行币都要将现金兑完,所以可以将拆解成两段不同的最短路,一条是从1点出发到i点代表用现金的最短路,直接dijkstra即可;另一条则是从i点到达n点的代表用旅行货币的最短路,反向建图再dijkstra即可。
再枚举每个点作为兑换点时从1~n所需要的现金res[i]并加入multise中,每次修改汇率就是将res[i]删掉,再重新计算res[i]加入到multise中,multise.begin()即为每次修改后的现金最小花费值。
但是我这里的代码还有个一分的测试点一直过不了qaq。

代码:

#include <bits/stdc++.h>
#define fastio ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL)
#define debug(a) cout << "debug : " << (#a) << " = " << a << endl

using namespace std;

typedef long long ll;
typedef pair<ll, int> PII;

const int N = 1e5 + 10;
const ll INF = 1e18;
const double eps = 1e-6;
const int mod = 998244353;

int n, m, k;
vector<PII> a[2][N];
int hui[N];
ll dis[2][N];
bool vis[N];

void dijkstra(int idx)
{
    memset(vis, false, sizeof vis);
    for (int i = 1; i <= n; i++)
        dis[idx][i] = INF;
    priority_queue<PII, vector<PII>, greater<PII>> pq;
    if (idx == 0)
    {
        pq.push({0, 1});
        dis[idx][1] = 0;
    }
    else
    {
        pq.push({0, n});
        dis[idx][n] = 0;
    }

    while (pq.size())
    {
        PII t = pq.top();
        pq.pop();
        int l = t.second;
        ll distance = t.first;
        for (auto p : a[idx][l])
        {
            int r = p.second;
            if (dis[idx][r] > distance + p.first)
            {
                dis[idx][r] = distance + p.first;
                pq.push({dis[idx][r], r});
            }
        }
    }
}

int main()
{
    fastio;
    cin >> n >> m >> k;
    while (m--)
    {
        int l, r, c, d;
        cin >> l >> r >> c >> d;
        a[0][l].push_back({c, r});
        a[1][r].push_back({d, l});
    }
    for (int i = 1; i <= n; i++)
        cin >> hui[i];
    dijkstra(0);
    dijkstra(1);
    multiset<ll> Set;
    ll res[N] = {0}; //res[i]为在i点换所要花费的
    for (int i = 1; i <= n; i++)
    {
        res[i] += dis[0][i];
        res[i] += (dis[1][i] - dis[1][n]) / hui[i];
        if (dis[1][i] % hui[i] != 0)
            res[i]++;
        Set.insert(res[i]);
    }
    while (k--)
    {
        int i, v;
        cin >> i >> v;
        if (res[i] != 0)
            Set.erase(Set.find(res[i]));
        hui[i] = v;
        res[i] = 0;
        res[i] += dis[0][i];
        res[i] += (dis[1][i] - dis[1][n]) / hui[i];
        if (dis[1][i] % hui[i] != 0)
            res[i]++;
        Set.insert(res[i]);
        cout << *Set.begin() << endl;
    }

    return 0;
}

标签:pq,idx,int,res,hui,L3,multiset,028,dis
来源: https://blog.csdn.net/Lucky12138/article/details/120544298