送外賣 （dfs）

原博客：https://blog.csdn.net/qq_39696016/article/details/104376916?depth_1-utm_source=distribute.pc_relevant.none-task&utm_source=distribute.pc_relevant.none-task

https://ac.nowcoder.com/acm/problem/13224

從初始位置開始遍歷，無非就是選a或b，題目要求選擇字典序最小，所以dfs的時候先a後b，

vis[pos] == 2 表示屬於剛剛走過，正在嘗試經過該位置看能不能字典序最小的解，或者Infinity!
vis[pos] == 1 表示屬於走過這個路程，並且找不到字典序最小的解，或者Infinity!

#include <iostream>
#include <bitset>
#include <algorithm>
#include <cmath>
#include <ctype.h>
#include <string>
#include <cstdio>
#include <set>
#include <cstdio>
#include <fstream>
#include <deque>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#include <iomanip>
#define SIS std::ios::sync_with_stdio(false)
#define ll long long
#define INF 0x3f3f3f3f
const int mod = 1e9 + 7;
const double esp = 1e-5;
const double PI = 3.141592653589793238462643383279;
using namespace std;
const int N = 1e7 + 5;
const int maxn = 1 << 20;
ll powmod(ll a, ll b) { ll res = 1; a %= mod; while (b >= 0); for (; b; b >>= 1) { if (b & 1)res = res * a % mod; a = a * a % mod; }return res; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
/*void chafen(int l, int r, int k) {//差分函數
	p[l] += k;
	p[r + 1] -= k;
}*/
/*************************************************************/	
ll a[N];
ll b[N];
char ans[N];//儲存答案
int c[N];
int vis[N],n;
int flag = 0;
bool dfs(int pos,int cur) {//cur表示走了幾次,pos表示當前到的位置
	if (pos<1 || pos>n) return false;
	if (pos == n) {
		ans[cur] = '\0';
		flag = 1;
		return true;
	}
	if (vis[pos] == 1) {
		return false;
	}
	else if (vis[pos]==2) {
		c[pos] = 1;
		return false;
	}
	vis[pos] = 2;//這個點剛走過
	ans[cur] = 'a';//字典序要求最小，從a先開始
	if (dfs(pos + a[pos], cur + 1)) {
		if (c[pos]) 
			flag = 2;
			return true;
		
	}
	ans[cur] = 'b';
	if (dfs(pos + b[pos], cur + 1)) {
		if (c[pos]) 
			flag = 2;
			return true;
		
	}
	vis[pos]--;
	return false;
}
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	for (int i = 1; i <= n; i++) {
		cin >> b[i];
	}
	dfs(1, 0);
	if (flag == 0) cout << "No solution!" << endl;
	else if (flag == 1) cout << ans << endl;
	else if (flag == 2) cout << "Infinity!" << endl;
	return 0;
}
/*
7
5 -3 6 5 -5 -1 6
-6 1 4 -2 0 -2 0
*/