178. 第K短路 题解

题目描述

avatar

解题思路

题解参考https://www.acwing.com/solution/content/21233/

代码

cpp
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
const int N = 1010, M = 20010;
typedef pair<int,int> PII;
typedef pair<int,PII> PIII;

int n, m, S, T, k;
int h[N], rh[N], e[M], w[M], ne[M], idx;
int dist[N], cnt[N];
bool vis[N];

void add(int h[], int a, int b, int c)
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

void dijkstra()  // 用dijkstra求出预估价值
{
    memset(dist, 0x3f, sizeof dist);
    priority_queue<PII, vector<PII>, greater<PII>> q;
    dist[T] = 0;
    q.push({0, T});

    while (q.size())
    {
        auto t = q.top();
        q.pop();

        int ver = t.y;
        if (vis[ver]) continue;
        vis[ver] = true;

        for (int i = rh[ver]; ~i; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > dist[ver] + w[i])
            {
                dist[j] = dist[ver] + w[i];
                q.push({dist[j], j});
            }
        } 
    }
}

int astar()
{
    priority_queue<PIII, vector<PIII>, greater<PIII>> q;
    q.push({dist[S],{0, S}});

    while (q.size())
    {
        auto t = q.top();
        q.pop();

        int distance = t.y.x, ver = t.y.y;
        cnt[ver]++;
        if (cnt[T] == k)    return distance;  //终点第k次出队,说明找到了第k短路

        for (int i = h[ver]; i != -1; i = ne[i])
        {
            int j = e[i];
            
            if (cnt[j] < k)
                q.push({dist[j] + distance + w[i],{distance + w[i], j}});
        }
    }
    return -1;
}

int main()
{
    memset(h, -1, sizeof h);
    memset(rh, -1, sizeof rh);

    scanf("%d%d", &n, &m);
    while (m--)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(h, a, b, c);
        add(rh, b, a, c);
    }

    scanf("%d%d%d", &S, &T, &k);
    if (S == T) k++;

    dijkstra();
    cout << astar();
    return 0;
}
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