Lift Hopping)

Lift Hopping)

Meaning: n elevators and 100 floors tell the speed of each elevator and the number of floors where each elevator will stop. the start point is at Layer 0, and the end point is at Layer k, in addition, you need to wait 60 seconds for each transfer and ask the shortest time from 0 to k.

Idea: This is to read the question, it hurts, and then run Dijkstra after creating a graph based on the input. Add 60 seconds to the update.

Code:

 

#include 
 
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   #include 
   
    #include #include 
    
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          #include 
          
           #pragma comment (linker,/STACK:102400000,102400000)#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b) for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i--)#define FRL(i,a,b) for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i--)#define mem(t, v) memset ((t) , v, sizeof(t))#define sf(n) scanf(%d, &n)#define sff(a,b) scanf(%d %d, &a, &b)#define sfff(a,b,c) scanf(%d %d %d, &a, &b, &c)#define pf printf#define DBG pf(Hi)typedef long long ll;using namespace std;#define INF 0x3f3f3f3f#define mod 1000000009const int maxn = 105;const int MAXN = 3005;const int MAXM = 200010;const int N = 1005;int mp[maxn][maxn],TIME[6];int dist[maxn];bool vis[maxn];int n,k;char str[MAXN];int a[maxn];int readLine(char *s){ int L; for (L=0;(s[L]=getchar())!=''&&s[L]!=EOF;L++) ; s[L]=0; return L;}void Dijkstra(){ int i,j,now,mi; mem(dist,INF); mem(vis,false); for (i=0;i<100;i++) dist[i]=mp[0][i]; dist[0]=0; vis[0]=true; for (i=0;i<100;i++) { now=-1; mi=INF; for (j=0;j<100;j++) { if (!vis[j]&&mi>dist[j]) { now=j; mi=dist[j]; } }// printf(now=%d,now); if (now==-1) break;// DBG; vis[now]=true; for (j=0;j<100;j++) { if (!vis[j]&&dist[j]>dist[now]+mp[now][j]+60) dist[j]=dist[now]+mp[now][j]+60; } }// for (i=0;i<=30;i++)// pf(%d ,dist[i]); if (dist[k]>=INF) pf(IMPOSSIBLE); else pf(%d,dist[k]);}int main(){#ifndef ONLINE_JUDGE freopen(C:/Users/lyf/Desktop/IN.txt,r,stdin);#endif int i,j; while (~sff(n,k)) { mem(mp,INF); for (i=0;i