题解

#include<bits/stdc++.h>
using namespace std;
const string days[]={
	"Today is Monday.",
	"Today is Tuesday.",
	"Today is Wednesday.",
	"Today is Thursday.",
	"Today is Friday.",
	"Today is Saturday.",
	"Today is Sunday."
};
map<string,int>num;
string name[22];
int n,m,p,cnt,ans,TF[102],T,F;
struct word{
	int id;
	string st;
}f[102];
string s;
bool pdTF(int id,bool b){
	if(TF[id]==-1){
		TF[id]=b;
		if(b)++T;else ++F;
	}else{
		if(TF[id]==b)return 0;else return 1;
	}
	if(F>m||T>n-m)return 1;
	return 0;
}
void Judge(int Xs,string day){
	memset(TF,-1,sizeof(TF));
	T=F=0;
	for(int i=1;i<=p;++i){
		int pos;
		pos=f[i].st.find("I am guilty.");
		if(~pos){
			if(pdTF(f[i].id,f[i].id==Xs))return;
		}
		pos=f[i].st.find("I am not guilty.");
		if(~pos){
			if(pdTF(f[i].id,f[i].id!=Xs))return;
		}
		pos=f[i].st.find(" is guilty.");
		if(~pos){
			string now=f[i].st;
			now.erase(pos,11);
			int id=num[now];
			if(pdTF(f[i].id,id==Xs))return;
		}
		pos=f[i].st.find(" is not guilty.");
		if(~pos){
			string now=f[i].st;
			now.erase(pos,15);
			int id=num[now];
			if(pdTF(f[i].id,id!=Xs))return;
		}
		pos=f[i].st.find("Today is ");
		if(~pos){
			if(pdTF(f[i].id,f[i].st==day))return;
		}
	}
	if(ans&&ans!=Xs){
		cout<<"Cannot Determine"<<endl;
		exit(0);
	}
	ans=Xs;
}
int main(){
	cin>>n>>m>>p;
	for(int i=1;i<=n;++i){
		cin>>name[i];
		num[name[i]]=i;
	}
	for(int i=1;i<=p;++i){
		cin>>s;
		s.erase(s.length()-1,1);
		f[i].id=num[s];
		getline(cin,f[i].st);
		f[i].st.erase(0,1);
		char ch=f[i].st[f[i].st.length()-1];
		if(ch=='\n'||ch=='\r'||ch==' ')
		f[i].st.erase(f[i].st.length()-1,1);
	}
	ans=0;
	for(int i=1;i<=n;++i)
	for(int j=0;j<7;++j)
	Judge(i,days[j]);
	if(!ans)
	cout<<"Impossible"<<endl;else
	cout<<name[ans]<<endl;
	return 0;
}

内存击败所有人
我们枚举犯人和星期，一个一个进行判断。如果成功则记录答案，如果成功且以前已经记录了答案，则说明有多个凶手，输出“Cannot Determine”，如果最后没有答案，则输出“Impossible”。主要是字符串的处理。