codeforces 689 Mike and Shortcuts(最短路)

codeforces 689 Mike and Shortcuts(最短路)

原题

1. 任意两点的距离是序号差，那么相邻点之间建边即可，同时加上题目提供的边
2. 跑一遍dijkstra可得1点到每个点的最短路，时间复杂度是O(mlogm)

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
const int Max=200000+10;
int abs(int x) {return x>=0?x:-x;}
int n,m;
struct Edge{
int from,to,dist;
Edge(int u,int v,int d):from(u),to(v),dist(d){}
};
vector <Edge> edges;
vector <int>  G[Max];
int d[Max];
void Init()
{
    for(int i=0;i<=n;i++) G[i].clear();
    edges.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);
}
struct node
{
    int d,u;
    node(int dd,int uu):d(dd),u(uu){};
    bool operator < (const node & rhs) const {
    return d>rhs.d;
    }
};
const int INF=0x3f3f3f3f;
bool done[Max];
void Dijkstra(int s)
{
    priority_queue<node>Q;
    for(int i=1;i<=n;i++) d[i]=INF;
    d[s]=0;
    memset(done,0,sizeof(done));
    Q.push(node(0,s));
    while(!Q.empty())
    {
        node x=Q.top();Q.pop();
        int u=x.u;
        if(done[u]) continue;
        done[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;
                Q.push(node(d[e.to],e.to));
            }
        }
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
       int x;
       Init();
       for(int i=1;i<=n;i++)
       {
           scanf("%d",&x);
           AddEdge(i-1,i,1);
           AddEdge(i+1,i,1);
           AddEdge(i,x,1);
       }
       Dijkstra(1);
       for(int i=1;i<=n;i++)
       {
           if(i!=1) printf(" ");
           printf("%d",d[i]);
       }
       puts("");
    }
    return 0;
}