题目描述

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;
}
