【luogu P4568 [JLOI2011]飞行路线】 题解

题目链接：https://www.luogu.org/problemnew/show/P4568
卡了一晚上，算是分层图最短路的模板。注意卡SPFA，所以我写了个SLF优化。
同时 AC400祭！~

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ri register
using namespace std;
const int maxn = 200000 + 10;
const int inf = 2147483647;
inline int read()
{
    int k=0,f=1;
    char c=getchar();
    while(!isdigit(c))
    {
        if(c=='-')f=-1;
        c=getchar();
    }
    while(isdigit(c))
    {
        k=(k<<1)+(k<<3)+c-48;
        c=getchar();
    }
    return k*f;
}
int n, m, k, s, end, dis[maxn][20];
bool vis[maxn][20];
struct edge{
    int from, len, to, next;
}e[maxn<<2];
int head[maxn], cnt = 0;
struct que{
    int a, b;
};
deque<que> q;
inline void add(int u, int v, int w)
{
    e[++cnt].from = u;
    e[cnt].to = v;
    e[cnt].len = w;
    e[cnt].next = head[u];
    head[u] = cnt;
}
inline void SPFA()
{
    memset(dis, 127, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    q.push_back((que){s,0});
    dis[s][0] = 0;
    vis[s][0] = 1;
    while(!q.empty())
    {
        que now = q.front(); q.pop_front();
        vis[now.a][now.b] = 0;
        for(ri int i = head[now.a]; i != -1; i = e[i].next)
        {
            if(dis[e[i].to][now.b] > dis[now.a][now.b] + e[i].len)
            {
                dis[e[i].to][now.b] = dis[now.a][now.b] + e[i].len;
                if(vis[e[i].to][now.b] == 0)
                {
                    vis[e[i].to][now.b] = 1;
                    if(q.empty() || dis[e[i].to][now.b] > dis[q.front().a][now.b])
                          q.push_back((que){e[i].to, now.b});
                    else
                    q.push_front((que){e[i].to, now.b});
                }
            }
            if(now.b + 1 <= k)
            {
                if(dis[e[i].to][now.b + 1] > dis[now.a][now.b])
                {
                    dis[e[i].to][now.b + 1] = dis[now.a][now.b];
                    if(vis[e[i].to][now.b + 1] == 0)
                    {
                            vis[e[i].to][now.b + 1] = 1;
                            if(q.empty() || dis[e[i].to][now.b] > dis[q.front().a][now.b + 1])
                        q.push_back((que){e[i].to, now.b + 1});
                        else
                        q.push_front((que){e[i].to, now.b + 1});
                    }
                }
            }
        }
    }
}
int main()
{
    //freopen(".in","r",stdin);
    //freopen(".out","w",stdout);
    memset(head, -1, sizeof(head));
    n = read(); m = read(); k = read(); s = read(); end = read();
    for(ri int i = 1; i <= m; i++)
    {
        int u, v, w;
        u = read(); v = read(); w = read();
        add(u, v, w); add(v, u, w);
    }
    SPFA();
    printf("%d",dis[end][k]);
    return 0;
}

原文地址：https://www.cnblogs.com/MisakaAzusa/p/9326154.html