• LUOGU P1039 侦探推理 (字符串+模拟)


    传送门

    解题思路

      一道%你神题,(string)好强大啊。。首先枚举一个周几,再枚举一个罪犯是谁,然后判断的时候就是枚举所有人说的话。定义(fAKe[i])表示第(i)个人说的是真话还是假话还是未知。然后如果遇到(xx is guilty)(I am guilty)之类的话,就看枚举的罪犯是否为这个人。如果说的是(Today is xxx),就看当前枚举的星期是否为这一天。

    代码

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<cstdlib>
    #include<string>
    #include<map>
    
    using namespace std;
    const int MAXN = 25;
    
    int n,m,p,fAKe[MAXN],w[205],ans,now,err;
    string name[MAXN],say[205];
    string day[10]={"","Today is Sunday.","Today is Monday.","Today is Tuesday.","Today is Wednesday."
                    ,"Today is Thursday.","Today is Friday.","Today is Saturday."};
    map<string,int> mp;
    
    void solve(int x,int y){
        if(!fAKe[x]) fAKe[x]=y;
        else if(fAKe[x] && fAKe[x]!=y) err=1;
    }
    
    int main(){
        scanf("%d%d%d",&m,&n,&p);
        for(int i=1;i<=m;i++) cin>>name[i],mp[name[i]]=i;
        for(int i=1;i<=p;i++){
            string nm;cin>>nm;
            nm.erase(nm.end()-1);w[i]=mp[nm];
            getline(cin,say[i]);say[i].erase(say[i].begin());
    		say[i].erase(say[i].end()-1);
        }
        for(int td=1;td<=7;td++)
            for(int chp=1;chp<=m;chp++){ //枚举罪犯 
                err=0;int who,cnt,pp;
                memset(fAKe,0,sizeof(fAKe));
                for(int i=1;i<=p;i++){
                    who=w[i];
                    if(say[i]=="I am guilty.") solve(who,who==chp?1:-1); 
                    if(say[i]=="I am not guilty.") solve(who,who==chp?-1:1);
                    for(int j=1;j<=7;j++) 
                        if(say[i]==day[j]) {solve(who,j==td?1:-1);break;}
                    for(int j=1;j<=m;j++){
                        if(say[i]==name[j]+" is guilty.") solve(who,chp==j?1:-1);
                        if(say[i]==name[j]+" is not guilty.") solve(who,chp==j?-1:1);
                    }		
                }
                cnt=pp=0;
                for(int i=1;i<=m;i++) {
                    if(!fAKe[i]) pp++;
                    if(fAKe[i]==-1) cnt++;
                }
                if(err || cnt>n || cnt+pp<n) continue;        
                if(ans && ans!=chp) {
                    puts("Cannot Determine");
                    return 0;
                }
                ans=chp;
            }
        if(ans) cout<<name[ans];
        else puts("Impossible");
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/sdfzsyq/p/9913015.html
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