CF1204C Anna, Svyatoslav and Maps

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 ;
} 

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