CodeChef Digit Jumps解题报告

Digit Jumps

题面

中文题面在这里：https://s3.amazonaws.com/codechef_shared/download/translated/JUNE14/mandarin/DIGJUMP.pdf
Chef loves games! But he likes to invent his own. Now he plays game "Digit Jump". Chef has sequence of digits S1, S2,..., SN,. He is staying in the first digit (S1) and want to reach the last digit (SN) in the minimal number of jumps.

While staying in some digit x with index i (digit Si) Chef can jump into digits with indices i - 1 (Si-1) and i + 1 (Si+1) but he can't jump out from sequence. Or he can jump into any digit with the same value x.

Help Chef to find the minimal number of jumps he need to reach digit SN from digit S1.

Input
Input contains a single line consist of string S of length N- the sequence of digits.

Output
In a single line print single integer - the minimal number of jumps he needs.

Constraints
1 ≤ N ≤ 10^5
Each symbol of S is a digit from 0 to 9.

Example
Input:
01234567890

Output:
1

Input:
012134444444443

Output:
4

Explanation

In the first case Chef can directly jump from the first digit (it is 0) to the last (as it is also 0).

In the second case Chef should jump in such sequence (the number of digits from 1: 1-2-4-5-15).

题解

我看好多好多人都说这个题好简单的好像直接bfs就跳完了
是不是我也可以这么说？？

代码（蒯的）

#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;
#define Maxn 100000
char s[Maxn+5];
int n;
int a[Maxn+5];
queue<int> q;
vector<int> x[10];
bool in[Maxn+5];
bool vis[10];
void bfs(){
    q.push(1);
    a[1]=0;
    in[1]=1;
    int now;
    while(!q.empty()){
        now=q.front();
        if(now==n){
            break;
        }
        in[now]=0;
        q.pop();
        if(now-1>0&&a[now-1]>a[now]+1){
            a[now-1]=a[now]+1;
            if(!in[now-1]){
                q.push(now-1);
                in[now-1]=1;
            }
        }
        if(now+1<=n&&a[now+1]>a[now]+1){
            a[now+1]=a[now]+1;
            if(!in[now+1]){
                q.push(now+1);
                in[now+1]=1;
            }
        }
        if(vis[s[now]-'0']){
            continue;
        }//每一个数只要跳一次即可，后面来的肯定不会更优
        vis[s[now]-'0']=1;
        for(int i=0;i<(int)x[s[now]-'0'].size();i++){
            if(a[x[s[now]-'0'][i]]>a[now]+1){
                a[x[s[now]-'0'][i]]=a[now]+1;
                if(!in[x[s[now]-'0'][i]]){
                    q.push(x[s[now]-'0'][i]);
                    in[x[s[now]-'0'][i]]=1;
                }
            }
        }
    }
}
int main(){
    memset(a,0x3f,sizeof a);
    scanf("%s",s+1);
    while(s[++n]!=''){
        x[s[n]-'0'].push_back(n);
    }
    n--;
    bfs();
    printf("%d
",a[n]);
    return 0;
}