Topcoder 636 B [violent]

A arrange is provided, and some numbers are accidentally erased. But there are N element pairs that know that this sequence meets the requirements of I <J & A [I] <A [J, then let you restore this arrangement. Then, there are several arrays that meet the conditions.

Analysis: It is very simple and violent. Fill in the erased value in the brute force mode, then check the number of elements, and compare them.

Code:

#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <fstream>#include <numeric>#include <iomanip>#include <bitset>#include <list>#include <stdexcept>#include <functional>#include <utility>#include <ctime>using namespace std;#define PB push_back#define MP make_pair#define REP(i,n) for(i=0;i<(n);++i)#define FOR(i,l,h) for(i=(l);i<=(h);++i)#define FORD(i,h,l) for(i=(h);i>=(l);--i)typedef vector<int> VI;typedef vector<string> VS;typedef vector<double> VD;typedef long long LL;typedef pair<int,int> PII;class SortishDiv2{        public:        int ways(int sortedness, vector <int> seq)        {            int ans = 0;            int vis[200],dis[200];            int pps[200];            memset(vis,0,sizeof(vis));            memset(dis,0,sizeof(dis));            for(int i=0;i<seq.size();i++)            {                dis[seq[i]]=1;                if(seq[i]==0)                {                    vis[i]=1;                    continue;                }                for(int j=i+1; j<seq.size();j++)                {                    if(seq[i]==0)                        continue;                    if(seq[i]<seq[j])                        ans++;                }            }            sortedness -= ans;            int count = 0;            vector<int> cc,wei;            for(int i=1;i<=seq.size();i++){                if(dis[i]==0){                    cc.push_back(i);                }                if(vis[i-1])                    wei.push_back(i-1);            }            if(cc.size()==0)                return sortedness==0;            do            {                ans = 0;                for(int i=0;i<seq.size();i++)                    pps[i] = seq[i];                for(int i = 0;i<wei.size();i++)                {                    for(int k=0;k<seq.size();k++)                    {                        if(pps[k]==0)                            continue;                        if(k>wei[i] && pps[k]>cc[i] || k<wei[i] && pps[k]<cc[i]){                            ans++;                        }                    }                    pps[wei[i]]=cc[i];                }                if(ans == sortedness)                    count++;            }while(next_permutation(cc.begin(),cc.end()));            cc.clear();            wei.clear();            return count;        }};int main(){    freopen("Input.txt","r",stdin);        SortishDiv2 a;        int n;        vector<int> v;        while(~scanf("%d",&n))        {            for(int i=0;i<n;i++)            {                int x;                scanf("%d",&x);                v.push_back(x);            }            cout<<a.ways(n,v)<<endl;            v.clear();        }        return 0;}

Topcoder 636 B [violent]