Codeforces 202B Brand New Easy Problem（模拟）

题意：

让你在m个序列中看是否存在给定序列的一个排列，并且计算逆序对数最大值。在找的时候我已开始想通过状压来枚举所有的排列，但是发现效率太慢了，于是听了队友的建议，对于给定序列的每一个排列，对所有的搜索串中进行搜索，这样就快多了。

代码：

////  Created by  CQU_CST_WuErli//  Copyright (c) 2015 CQU_CST_WuErli. All rights reserved.//// #include<bits/stdc++.h>#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <string>#include <vector>#include <list>#include <map>#include <queue>#include <stack>#include <set>#include <algorithm>#include <sstream>#define CLR(x) memset(x,0,sizeof(x))#define OFF(x) memset(x,-1,sizeof(x))#define MEM(x,a) memset((x),(a),sizeof(x))#define For_UVa if (kase!=1) cout << endl#define BUG cout << "I am here" << endl#define lookln(x) cout << #x << "=" << x << endl#define SI(a) scanf("%d",&a)#define SII(a,b) scanf("%d%d",&a,&b)#define SIII(a,b,c) scanf("%d%d%d",&a,&b,&c)#define rep(flag,start,end) for(int flag=start;flag<=end;flag++)#define Rep(flag,start,end) for(int flag=start;flag>=end;flag--)#define Lson l,mid,rt<<1#define Rson mid+1,r,rt<<1|1#define Root 1,n,1#define BigInteger bigntemplate <typename T> T gcd(const T& a,const T& b) {return b==0?a:gcd(b,a%b);}const int MAX_L=2005;// For BigIntegerconst int INF_INT=0x3f3f3f3f;const long long INF_LL=0x7fffffff;const int MOD=1e9+7;const double eps=1e-9;const double pi=acos(-1);typedef long long  ll;using namespace std;int n;int m;vector<int> pro;vector<int> arc[11];map<string,int> mp;int idx;void init() {    pro.clear();    for (int i=0;i<11;i++) arc[i].clear();    mp.clear();    idx=1;}int get(string& s) {    if (!mp.count(s)) mp[s]=idx++;    return mp[s];}void print() {    for (int i=0;i<pro.size();i++) cout << pro[i] << ' ';    cout << endl;    for (int i=1;i<=m;i++) {        for (int j=0;j<arc[i].size();j++) cout << arc[i][j] << ' ';        cout << endl;    }}int cal() {    map<int,int> mpx;    int ret=0;    for (int i=0;i<n;i++) {        mpx[pro[i]]++;        for (int j=pro[i]+1;j<=n;j++) ret+=mpx[j];    }    return (n*(n-1)/2)-ret+1;}bool check(int x) {    int cnt=0;    for (int i=0;i<arc[x].size();i++) {        if (arc[x][i]==pro[cnt]) {            cnt++;        }        if (cnt==n) break;    }    return cnt==n;}int main(){#ifdef LOCAL    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);//  freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);#endif    while (SI(n)==1) {        init();        for (int i=1;i<=n;i++) {            string x;            cin >> x;            pro.push_back(get(x));        }        SI(m);        for (int i=1;i<=m;i++) {            int x;SI(x);            for (int j=1;j<=x;j++) {                string tmp;                cin >> tmp;                arc[i].push_back(get(tmp));            }        }//      print();        int ans=0,id=-1;        do {            int tmp=cal();            int ok=0;            int i;            for (i=1;i<=m;i++) if (check(i)) {                ok=1;                break;            }            if (ok) {//              cout << "i=" << i << ' ' << "tmp=" << tmp << endl;                if (ans<tmp){                    ans=tmp;                    id=i;                }                if (ans==tmp && id>i) id=i;            }        }while (next_permutation(pro.begin(),pro.end()));        if (id==-1) cout << "Brand new problem!" << endl;        else {            cout << id << endl;            cout << "[:";            for (int i=1;i<=ans;i++) cout << '|';            cout << ":]\n";        }    }    return 0;}