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

原文地址：https://www.cnblogs.com/lyt020321/p/11391125.html