Topcoder srm631 div2 solution report

250: the grid has two colors. The maximum value of the longest continuous color in a column in the grid is the same.

Solution: violence.

Solution code:

// BEGIN CUT HERE/**/// END CUT HERE#line 7 "TaroGrid.cpp"#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 TaroGrid{        public:        int getNumber(vector <string> grid)        {             int n = grid[0].size();             int mx = 1;              int len = grid.size();             for(int i = 0 ;i < n ;i++)             {               int temp = grid[0][i];                int sum = 1 ;                for(int j = 1;j < len ;j ++)               {                 if(grid[j][i] == temp)                 {                    sum ++ ;                  }else {                    sum = 1 ;                    temp = grid[j][i];                 }                 if(sum > mx)                     mx = sum ;               }             }          return mx;        }        };

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500, ask if you can make each location have at most one cat.

Solution: sort the vertices of a cat and put all the cats on the left as much as possible. If the cat cannot be placed on the left or there is no repeating interval, it is impossible.

Solution code:

 1 // BEGIN CUT HERE 2 /* 3  4 */ 5 // END CUT HERE 6 #line 7 "CatsOnTheLineDiv2.cpp" 7 #include <cstdlib> 8 #include <cctype> 9 #include <cstring>10 #include <cstdio>11 #include <cmath>12 #include <algorithm>13 #include <vector>14 #include <string>15 #include <iostream>16 #include <sstream>17 #include <map>18 #include <set>19 #include <queue>20 #include <stack>21 #include <fstream>22 #include <numeric>23 #include <iomanip>24 #include <bitset>25 #include <list>26 #include <stdexcept>27 #include <functional>28 #include <utility>29 using namespace std;30 31 #define PB push_back32 #define MP make_pair33 34 #define REP(i,n) for(i=0;i<(n);++i)35 #define FOR(i,l,h) for(i=(l);i<=(h);++i)36 #define FORD(i,h,l) for(i=(h);i>=(l);--i)37 38 typedef vector<int> VI;39 typedef vector<string> VS;40 typedef vector<double> VD;41 typedef long long LL;42 typedef pair<int,int> PII;43 44 struct node{45   int x, y; 46 }cat[100];47 bool cmp(node a, node b )48 {49    return a.x < b.x;50 }51 class CatsOnTheLineDiv252 {53         public:54         string getAnswer(vector <int> p, vector <int> c, int t)55         {56            int n = p.size();57            for(int i = 0 ; i< n;i ++ )58            {59               cat[i].x = p[i];60               cat[i].y = c[i];61            }62            sort(cat,cat+n,cmp);63            int s = -1e9;64            int ok = 1;65            for(int i = 0 ;i < n;i ++)66            {67         68              int k1 = max(s+1,cat[i].x-t);69              int k2 = k1 + cat[i].y -1;70              //printf("%d %d\n",k1,k2);71              if(k2 - cat[i].x > t)72                  ok  = 0 ; 73              s = k2;74            }75            if(!ok)76                return "Impossible";77            else return "Possible";78           79         }80         81 82 };

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Topcoder srm631 div2 solution report