CF1204C Anna, Svyatoslav and Maps

Posted 2022-08-12 lyt020321

题意

在给定的序列P中求一个子序列,使得在图中按照该子序列进行最短路径移动时可以完整经过原序列P

code

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 105
#define maxm 1000010
#define inf 0x3f3f3f3f
using namespace std ;
int n ,m , idx ;
char mp[maxn][maxn] ;
int G[maxn][maxn] , point[maxm] , ans[maxm] ;
int qu[maxm*2] ;
int head = 1 , tail = 0 ; 
int main () 
    memset(G,0x3f,sizeof(G)) ;
    cin >> n ;
    for(int i = 1 ; i <= n ; i ++) 
        scanf("%s",mp[i]+1) ;
    
    for(int i = 1 ; i <= n ; i ++) 
        for(int j = 1 ; j <= n ; j ++) 
            if(mp[i][j] == '1') 
                G[i][j] = 1 ;
            
        
        G[i][i] = 1 ;
    
    for(int k = 1 ; k <= n ; k ++) 
        for(int i = 1 ; i <= n ; i ++) 
            for(int j = 1 ; j <= n ; j ++) 
                G[i][j] = min(G[i][j],G[i][k]+G[k][j]) ;
            
        
    
    cin >> m ;
    for(int i = 1 ; i <= m ; i ++) 
        cin >> point[i] ;
    
    int st = 1 , now = 2  ;
    while(now <= m) 
        int diss = now - st ;
        if(diss == G[point[st]][point[now]]) 
            if(head <= tail) 
                head ++ ;
            
            qu[++tail] = now ;
            now ++ ;
        else 
            ans[++idx] = point[st] ;
            if(head <= tail) 
                st = qu[head++] ;
            
        
    
    ans[++idx] = point[st] ;
    if(ans[idx] != point[m]) 
        ans[++idx] = point[m] ;
    
    cout << idx << endl ;
    for(int i = 1 ; i <= idx ; i ++) 
        cout << ans[i] << " " ;
    
    return 0 ;
 

溜了溜了