codeforces 1277 D xg
题意:给n个01字符串,让你进行k下操作,使得t字符串翻转,使得n个字符串存在一个顺序,使得ai+1的首字母和ai相等。使得k最小,输出这k个数。(注意变化前和变化后的字符串不重复)
思路:只有四种字符串,记录下每种字符串的数量。如果11,00,都不为0,且10,01为0,则无解。
考虑sum1和sum2的差值小于1是有解的,否则考虑使得数量多的。
若10多于01,考虑10的字符串翻转状态不存在在原字符串的数量够不够(sum10-sum01)/2。
字符串标记map就行。
#include <iostream> #include <cmath> #include <cstdio> #include <cstring> #include <string> #include <map> #include <iomanip> #include <algorithm> #include <queue> #include <stack> #include <set> #include <vector> // #include <bits/stdc++.h> #define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0); #define sp ' ' #define endl '\n' #define inf 0x3f3f3f3f; #define FOR(i,a,b) for( int i = a;i <= b;++i) #define bug cout<<"--------------"<<endl #define P pair<int, int> #define fi first #define se second #define pb(x) push_back(x) #define ppb() pop_back() #define mp(a,b) make_pair(a,b) #define ms(v,x) memset(v,x,sizeof(v)) #define rep(i,a,b) for(int i=a;i<=b;i++) #define repd(i,a,b) for(int i=a;i>=b;i--) #define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c)) #define sca2(a,b) scanf("%d %d",&(a),&(b)) #define sca(a) scanf("%d",&(a)); #define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c)) #define sca2ll(a,b) scanf("%lld %lld",&(a),&(b)) #define scall(a) scanf("%lld",&(a)); using namespace std; typedef long long ll; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;} const double Pi = acos(-1.0); const double epsilon = Pi/180.0; const int maxn = 2e5+10; map<string,int>vis; string a[maxn]; int ans[maxn]; int main() { // freopen("input.txt", "r", stdin); int _; scanf("%d",&_); while(_--) { int n; cin>>n; int tail = 0; int sum1 = 0,sum2 = 0,sum11=0,sum00=0; queue<int>que1; queue<int>que2; rep(i,1,n){ string s; cin>>s; a[i] = s; vis[s] = 1; int len = s.length()-1; if(s[0]=='0'&&s[len]=='0') sum00++; else if(s[0]=='0'&&s[len]=='1') { que2.push(i); sum2++; } else if(s[0]=='1'&&s[len]=='1') sum11++; else if(s[0]=='1'&&s[len]=='0') { que1.push(i); sum1++; } } if(sum11&&sum00&&sum1==0&&sum2==0){ cout<<-1<<endl; continue; } else { if(sum1 >= sum2){ int cha = sum1 - sum2; if (cha <= 1) { cout<<0<<endl; /* code */ } else { cha /= 2; tail = 0; while(que1.size()){ if(tail == cha) break; int pos = que1.front(); que1.pop(); string tmp = a[pos]; //int k = tmp.length(); reverse(tmp.begin(),tmp.end()); if(vis[tmp]==1) continue; ans[++tail] = pos; } if(tail == cha){ cout<<tail<<endl; rep(i,1,tail){ cout<<ans[i]<<sp; } cout<<endl; } else cout<<-1<<endl; } } else { int cha = sum2 - sum1; if (cha <= 1) { cout<<0<<endl; /* code */ } else { cha /= 2; tail = 0; while(que2.size()){ if(tail == cha) break; int pos = que2.front(); que2.pop(); string tmp = a[pos]; reverse(tmp.begin(),tmp.end()); if(vis[tmp]==1) continue; ans[++tail] = pos; } if(tail == cha){ cout<<tail<<endl; rep(i,1,tail){ cout<<ans[i]<<sp; } cout<<endl; } else cout<<-1<<endl; }cout<<endl; } } rep(i,1,n){ vis[a[i]] = 0; } //cout<<sum1<<sp<<sum2<<endl; } }