CDOJ1085 基爷与加法等式 爆搜DFS

题目链接:

http://acm.uestc.edu.cn/#/problem/show/1085

题意:

题解:

因为进位，从低位搜

代码:

http://www.cnblogs.com/qscqesze/p/4489781.html

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 1e5+10;

int n,ma,tmp,ans;
string s[30];
char k[100];
int tp[30],ans1[30],ans2[30],H[30],mp[30],vis[10];

bool check(){
    ans1[ma] = 0;
    for(int i=ma-1; i>=0; i--){
        ans1[i] = 0;
        ans2[i] = 0;
        for(int j=0; j<n-1; j++){
            if(H[s[j][i]-'A'] == -1) return 0;
            ans1[i] += H[s[j][i]-'A'];
        }
        if(H[s[n-1][i]-'A'] == -1) return 0;
        ans2[i] = H[s[n-1][i]-'A'];
        ans1[i] += (ans1[i+1]/10);
        if((ans1[i])%10 != ans2[i]) return 1;
    }

    if(ans1[0] > 9) return 1;
    return 0;
}

void dfs(int x){
    if(x==tmp) {
        ans++;
        return ;
    }

    for(int i=0; i<=9; i++){
        if(i==0 && tp[k[x]-'A']) continue;
        if(vis[i]) continue;
        vis[i] = 1;
        H[k[x]-'A'] = i;
        if(!check()) dfs(x+1);
        H[k[x]-'A'] = -1;
        vis[i] = 0;
    }

}

int main(){
    while(cin >> n){
        for(int i=0;i<30;i++)
            mp[i]=tp[i]=H[i]=0;
        ma=0;
        for(int i=0; i<n; i++){
            cin >> s[i];
            ma = max(ma,(int)s[i].size());
        }

        for(int i=0; i<n; i++)
            tp[s[i][0]-'A'] = 1;

        for(int i=0; i<n; i++){
            int len = ma-s[i].size();
            for(int j=0; j<len; j++){
                s[i] = '[' + s[i];
            }
        }

        tmp=0;
        ans=0;
        for(int i=ma-1;i>=0;i--){
            for(int j=0;j<n;j++){
                if(mp[s[j][i]-'A']) continue;
                if(s[j][i]=='[') continue;
                k[tmp++]=s[j][i];
                mp[s[j][i]-'A']=1;
            }
        }

        for(int i=0; i<tmp; i++)
            H[k[i]-'A'] = -1;

        // H['['-'A'] = 0;
        dfs(0);

        cout << ans << endl;
    }

    return 0;
}