題意:給出一個有向圖,起點終點以及k,求k短路
思路:使用A*算法,首先預處理出每個點到終點的距離h[i],搜索時使用優先隊列,關鍵字爲(f + h[i]),其中f 是已經走的距離,i是當前走到的節點
注意:此題坑點巨多:
1)起點可能與終點相同,此時0不算最短路,故k++
2)k短路不存在時,輸出-1
3)由於是有向圖,從Dijkstra時只能用反邊,A*時只能用正邊,否則MLE
代碼:
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
#define For(i,j,k) for(int i = j;i <= k;i ++)
#define Set(A, val) memset(A, val, sizeof(A))
const int N = 1010;
using namespace std;
struct Node{
int id, w;
bool operator < (const Node& a)const{
return w > a.w;
}
};
int d[N], vis[N], n, m;
vector<int> G[N], w[N];
vector<bool> P[N];
void Add(int x, int y, int c, int p){
G[x].push_back(y);
w[x].push_back(c);
P[x].push_back(p);
}
void Dijkstra(int st){
For(i,1,n) vis[i] = 0, d[i] = 1e9;
d[st] = 0;
priority_queue<Node> q;
q.push((Node){st, 0});
while(!q.empty()){
int s = q.top().id, g = q.top().w;
q.pop();
if(vis[s])continue;
vis[s] = 1;
For(i,0,(int)G[s].size() - 1){
int v = G[s][i];
if(!P[s][i] && d[v] > g + w[s][i])
d[v] = g + w[s][i], q.push((Node){v, d[v]});
}
}
}
struct Data{
int pos, dis;
bool operator < (const Data &a) const{
return dis + d[pos] > a.dis + d[a.pos];
}
};
int A_Star(int k, int st, int ed){
Set(vis, 0);
priority_queue<Data> q;
q.push((Data){st, 0});
while(!q.empty()){
Data s = q.top();q.pop();
int u = s.pos;
vis[u]++;
if(vis[ed] == k) return s.dis;
if(vis[u] > k) continue;
For(i,0,(int)G[u].size() - 1){
int v = G[u][i];
if(P[u][i])
q.push((Data){v, s.dis + w[u][i]});
}
}
return -1;
}
int main(){
int st, ed, k;
scanf("%d%d", &n, &m);
For(i,1,m){
int x, y, c;
scanf("%d%d%d", &x, &y, &c);
Add(x, y, c, 1);
Add(y, x, c, 0);
}
scanf("%d%d%d", &st, &ed, &k);
if(st == ed) ++k;
Dijkstra(ed);
printf("%d\n", A_Star(k, st, ed));
return 0;
}