Uva10067 playing with wheels [drawing + Shortest Path]

Question: uva10067 playing with wheels

Question: Give a machine with four cycles. See the figure below. Then give an initial number and a target number, and then a number that cannot appear during the period. You can change the number every minute, ask you the minimum required time.

Analysis: This question can be converted into the Shortest Path of the graph.

Because there are eight States that can be converted to the current state, we can convert each state to a vertex, and then the State is connected to an edge with a length of 1, find the shortest path from the initial state to the target State.

At the beginning, we created a graph for each set of data for 0.9 s, and optimized it for a moment, that is, we restored it after deleting the edge that cannot be reached for each graph creation, in this way, you don't need to create a graph every time. It takes 0.19 s and the time is quite fast. It's just a spfa time complexity O (m ),

AC code:

#include <cstdio>#include <iostream>#include <queue>#include <cstring>#include <algorithm>#include <stack>#include <vector>#include <utility>#include <cmath>using namespace std;const int N = 10005;const int M = 10000;const int inf = 0x3f3f3f3f;struct Node{    int x,len;};vector<Node> v[N];void add_Node(int x,int y,int len){    v[x].push_back((Node){y,len});    v[y].push_back((Node){x,len});}int count(int i,int j,int k,int f){    return ((i+10)%10)*1000+((j+10)%10)*100+((k+10)%10)*10+(f+10)%10;}int dir[10][5]={    {0,0,0,1},{0,0,0,-1},    {0,0,1,0},{0,0,-1,0},    {0,1,0,0},{0,-1,0,0},    {1,0,0,0},{-1,0,0,0}};void build(int i,int j,int k,int f){    int tmp=count(i,j,k,f);    for(int p=0; p<8; p++)    {        int tmp1=count(i+dir[p][0],j+dir[p][1],k+dir[p][2],f+dir[p][3]);        add_Node(tmp,tmp1,1);    }}void isit(){    for(int i=0; i<10; i++)        for(int j=0; j<10; j++)            for(int k=0; k<10; k++)                for(int f=0; f<10; f++)                    build(i,j,k,f);}int dis[N];void spfa(int s)  {    int i;    queue<int> q;    for(i=0; i<N; i++)        dis[i]=inf;    dis[s]=0;    q.push(s);    while(!q.empty())    {        int u=q.front();        q.pop();        for(i=0; i<v[u].size(); i++)        {            Node p=v[u][i];            if(dis[p.x]>dis[u]+p.len)            {                dis[p.x]=dis[u]+p.len;                q.push(p.x);            }        }    }}int dx[N],dy[N],dz[N],dk[N];int main(){    int T;    isit();    scanf("%d",&T);    while(T--)    {        int x,y,z,k;        scanf("%d%d%d%d",&x,&y,&z,&k);        int st=x*1000+y*100+z*10+k;        scanf("%d%d%d%d",&x,&y,&z,&k);        int en=x*1000+y*100+z*10+k;        int cc;        scanf("%d",&cc);        for(int i=0;i<cc;i++)        {            scanf("%d%d%d%d",&dx[i],&dy[i],&dz[i],&dk[i]);            int tmp=count(dx[i],dy[i],dz[i],dk[i]);            v[tmp].clear();        }        spfa(st);        if(dis[en]>=inf)            puts("-1");        else            printf("%d\n",dis[en]);        for(int i=0;i<cc;i++)        {            build(dx[i],dy[i],dz[i],dk[i]);        }    }}

Uva10067 playing with wheels [drawing + Shortest Path]