2018 TCO Algorithm Round 1B

250 LineOff

随便搞

#include <bits/stdc++.h>
using namespace std;

class LineOff {
public:
    int movesToDo(string points) {
        stack<char> st;
        int ret = 0, l = points.length();
        for(int i = 0; i < l; ++i){
            if(!st.empty() && st.top() == points[i]) ret++, st.pop();
            else st.push(points[i]);
        }
        return ret;
    }
};

600 StablePairsDiv1

随便搞

#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;
int f[111][111][111];
pii pre[111][111][111];

class StablePairsDiv1 {
public:
    void get(int step, int i, int j, vector<int> & ret){
        if(step != 1) get(step - 1, pre[step][i][j].first, pre[step][i][j].second, ret);
        ret.push_back(i);
        ret.push_back(j);
    }
    vector<int> findMaxStablePairs(int n, int c, int k) {
        memset(f, -1, sizeof(f));
        for(int i = 1; i <= n; ++i)
            for(int j = i + 1; j <= n; ++j)
                f[1][i][j] = i + j;
        for(int i = 1; i < k; ++i){
            for(int j = 1; j <= n; ++j){
                for(int k = j + 1; k <= n; ++k){
                    if(f[i][j][k] == -1) continue;
                    for(int p = k + 1; p <= n; ++p){
                        int q = j + k + c - p;
                        if(q <= p) break;
                        int t = f[i][j][k] + p + q;
                        if(t > f[i+1][p][q]) f[i+1][p][q] = t, pre[i+1][p][q] = pii(j, k);
                    }
                }
            }
        }
        int ans = -1, ii, jj;
        for(int i = 1; i <= n; ++i){
            for(int j = i + 1; j <= n; ++j){
                if(f[k][i][j] > ans) ans = f[k][i][j], ii = i, jj = j;
            }
        }
        vector<int> ret;
        if(ans == -1) return ret;
        get(k, ii, jj, ret);
        return ret;
    }
};

1000 ThreeSameLetters

随便搞

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod = 1e9 + 7;

LL f[66][66][4][2];
class ThreeSameLetters {
public:
    int countStrings(int L, int S){
        for(int i = 1; i <= S; ++i) f[1][i][1][0] = 1;
        for(int i = 1; i < L; ++i){
            for(int j = 1; j <= S; ++j){
                for(int k = 1; k <= 3; ++k){
                    for(int p = 0; p <= 1; ++p){
                        for(int q = 1; q <= S; ++q){
                            int nk = q == j ? k + 1 : 1;
                            if(nk > 3) continue;
                            if(p && nk == 3) continue;
                            int np = p || nk == 3;
                            f[i+1][q][nk][np] = (f[i+1][q][nk][np] + f[i][j][k][p]) % mod;
                        }
                    }
                }
            }
        }
        LL ans = 0;
        for(int i = 1; i <= S; ++i) ans = (ans + f[L][i][1][1] + f[L][i][2][1] + f[L][i][3][1]) % mod;
        return (int) ans;
    }
};