给定一个n个点,m条边的有向图
求s到t的第k短路
Input
第一行n和m(1<=n<=1000,1<=m<=100000)
接下来m行每行三个数a,b,t (1<=a,b<=n, 1<=t<=100) 表示有一条从a到b的有向边长为t
最后一行s,t,k
Output
一行一个整数,s到t 的第k短路,若不存在,则输出-1
Sample Input
2 2
1 2 5
2 1 4
1 2 2
Sample Output
14
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN = 1e5+5;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
struct Edge
{
int from, to; LL dist; //起点,终点,距离
Edge(int from, int to, LL dist):from(from), to(to), dist(dist) {}
};
struct node{
int f, g, from;
bool operator < (node a)const
{
if(a.f == f) {
return a.g < g;
}
return a.f < f;
}
};
struct Dijkstra
{
int n, m; //结点数,边数(包括反向弧)
vector<Edge> edges; //边表。edges[e]和edges[e^1]互为反向弧
vector<int> G[MAXN]; //邻接表,G[i][j]表示结点i的第j条边在edges数组中的序号
int vis[MAXN]; //标记数组
LL d[MAXN]; //s到各个点的最短路
int p[MAXN]; //上一条弧
vector<Edge>edges1;
vector<int>G1[MAXN];
void init(int n)
{
this->n = n;
edges.clear();
for (int i = 0; i <= n; i++) G[i].clear();
}
void AddEdge(int from, int to, int dist)
{
edges.push_back(Edge(from, to, dist));
m = edges.size();
G[from].push_back(m - 1);//反向建图dij用反向图
edges1.push_back(Edge(to, from, dist));
m = edges1.size();
G1[to].push_back(m - 1);//正向见图astart用正向图
}
struct HeapNode
{
int from; LL dist;
bool operator < (const HeapNode& rhs) const
{
return rhs.dist < dist;
}
HeapNode(int u, LL w): from(u), dist(w) {}
};
void dijkstra(int s)
{
priority_queue<HeapNode> Q;
for (int i = 0; i <= n; i++) d[i] = INF;
memset(vis, 0, sizeof(vis));
d[s] = 0;
Q.push(HeapNode(s, 0));
while (!Q.empty())
{
HeapNode x = Q.top(); Q.pop();
int u = x.from;
if (vis[u]) continue;
vis[u] = true;
for (int i = 0; i < G[u].size(); i++)
{
Edge& e = edges[G[u][i]];
if (d[e.to] > d[u] + e.dist)
{
d[e.to] = d[u] + e.dist;
p[e.to] = G[u][i];
Q.push(HeapNode(e.to, d[e.to]));
}
}
}
}//dij刘汝佳
int Astar(int s, int t, int k) {
if(s == t) {
k++;
}
if(d[s] == INF) {
return -1;
}
priority_queue <node> q;
int cnt = 0;
node tmp, to;
tmp.from = s;
tmp.g = 0;
tmp.f = tmp.g + d[tmp.from];
q.push(tmp);
while(!q.empty()) {
tmp = q.top();
q.pop();
if(tmp.from == t) {
cnt++;
}
if(cnt == k) {
return tmp.g;
}
for(int j = 0;j < G1[tmp.from].size();++j) {
Edge& e = edges1[G1[tmp.from][j]];
to.from = e.to;
to.g = tmp.g + e.dist;
to.f = to.g + d[to.from];
q.push(to);
}
}
return -1;
}
}gao;
int main() {
int n ,m;
scanf("%d%d", &n, &m);
gao.init(n);
while(m--) {
int a, b, t;
scanf("%d%d%d", &a, &b, &t);
gao.AddEdge(b, a, t);//反向存边
}
int s, t, k;
scanf("%d%d%d", &s, &t, &k);
gao.dijkstra(t);//反向存图跑f【】函数值
int ans = gao.Astar(s, t, k);//astart找第K短路
printf("%d\n", ans);
return 0;
}