【洛谷 P4568】 [JLOI2011]飞行路线 （分层最短路）

题目链接

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\n", dis[t + k * n]);
    return 0;
}