分层图最短路。
把每个点拆成(k+1)个点,表示总共有(k+1)层。
然后每层正常连边,
若((u,v))有边,则把每一层的(u)和下一层的(v)、每一层的(v)和下一层的(u)连边。
然后跑最短路就行了,终点是最后一层的(t)。
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
int xjc; char ch;
inline int read(){
xjc = 0; ch = getchar();
while(ch < '0' || ch > '9') ch = getchar();
while(ch >= '0' && ch <= '9'){ xjc = xjc * 10 + ch - '0'; ch = getchar(); }
return xjc;
}
const int MAXN = 100010;
const int MAXM = 500010;
struct Edge{
int next, to, dis;
}e[MAXM << 2];
int head[MAXN], num, dis[MAXN];
inline void Add(int from, int to, int dis){
e[++num] = (Edge){ head[from], to, dis }; head[from] = num;
}
typedef pair<int, int> point;
priority_queue <point, vector<point>, greater<point> > q;
point now;
int n, m, k, s, t, a, b, c;
int main(){
n = read(); m = read(); k = read(); s = read(); t = read();
for(int i = 1; i <= m; ++i){
a = read(); b = read(); c = read();
Add(a, b, c); Add(b, a, c);
for(int j = 1; j <= k; ++j){ //分层
Add(a + (j - 1) * n, b + j * n, 0);
Add(b + (j - 1) * n, a + j * n, 0);
Add(a + j * n, b + j * n, c);
Add(b + j * n, a + j * n, c);
}
}
q.push(point(0, s));
memset(dis, 127, sizeof dis);
dis[s] = 0;
while(q.size()){
now = q.top(); q.pop();
if(dis[now.second] < now.first) continue;
for(int i = head[now.second]; i; i = e[i].next)
if(dis[e[i].to] > dis[now.second] + e[i].dis){
dis[e[i].to] = dis[now.second] + e[i].dis;
q.push(point(dis[e[i].to], e[i].to));
}
}
printf("%d
", dis[t + k * n]);
return 0;
}