Topcoder SRM 630div 2

A: eliminate two adjacent equal characters.

I really don't like STL ..

#include<iostream>#include <string>#include <vector>using namespace std;class DoubleLetter {    public:    string ableToSolve(string S) {    while (1){    int flag=0;    for (int i=1;i<S.size();i++)    if (S[i]==S[i-1])    {       S.erase(i-1,2);       flag=1;    }    if (!flag||S.size()==0) break;    }    if (S.size()==0) return "Possible";    else return "Impossible";    }};// Powered by FileEdit// Powered by TZTester 1.01 [25-Feb-2003]// Powered by CodeProcessor

B: Give a tree with N nodes, and find the maximum m value (n <= 10) at the same distance between the points in m pairs );

First, Floyd finds the distance between all vertices.

Then, the solution is enumerated in a way similar to the State compressed DP.

Then compare the maximum value.

#include<iostream>#include <string>#include <vector>#include<string.h>using namespace std;int b[12345];int dis[12][12];int mp[12][12];class Egalitarianism3Easy {    public:    int maxCities(int n, vector <int> a, vector <int> b, vector <int> len) {    memset(dis,0x3f3f3f,sizeof(dis));        for (int i=1;i<=n;i++) dis[i][i]=0;    for (int i=0;i<a.size();i++) dis[a[i]][b[i]]=dis[b[i]][a[i]]=len[i];        for (int k=1;k<=n;k++)        for (int i=1;i<=n;i++)        for (int j=1;j<=n;j++)        dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);                int Max=1;        for (int i=1;i<(1<<n);i++)        {            vector<int> V;            V.clear();            for (int j=0;j<n;j++)            if (((i>>j)&1)==1) V.push_back(j+1);                        int m=V.size();            int same=dis[V[0]][V[m-1]];            int flag=1;                       for (int k=0;k<m;k++)                for (int j=0;j<m;j++)                if (k!=j&&dis[V[k]][V[j]]!=same)                {                flag=0;                break;            }           if (flag) Max=max(Max,m);        }        return Max;    }};

C: Exercise your thinking.

Reference others' practices:

Use an array similar to a suffix.

It's just that the array sets are really drunk.

My solution is to find the minimum string that satisfies the same sequence number as s, and then compare it with the value of S.

Detailed comments are provided in the Code.

# Include <iostream> # include <string> # include <vector> # include <string. h ># include <algorithm> using namespace STD; Class suffixarraydiv2 {public: String arr [100]; int P [100], a [100]; // note p: indicates the subscript of the original array after the string is sorted. // A: indicates the subscript of the original string after sorting. // This is very round, char C [100]; string smallerone (string s) {for (INT I = 0; I <S. size (); I ++) Arr [I] = & S [I]; // Save the suffix sort (ARR, arr + S. size (); int Len = S. size (); For (INT I = 0; I <S. size (); I ++) {P [I] = len-Arr [I]. size (); A [Len-Arr [I]. size ()] = I;} A [Len] =-1; C [p [0] = 'a'; For (INT I = 1; I <Len; I ++) {if (a [p [I-1] + 1] <A [p [I] + 1]) c [p [I] = C [p [I-1]; // This is a character we determine to position P [I], // if the current string S1, and the following string S2 (both sorted strings) // compare the size of the strings plus 1. Here I can ignore my explanation else C [p [I] = C [p [I-1] + 1;} string ans; For (INT I = 0; I <Len; I ++) ans + = C [I]; If (ANS <s) Return "exists"; else return "does not exist" ;}}; int main () {suffixarraydiv2 B; string s; CIN> S; cout <B. smallerone (s); Return 0 ;}

 

Topcoder SRM 630div 2