Topcoder SRM 648 Div.2 - A(水), B（枚举）,C（DP/构造）

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第一场TC~~ 一道题0.0 Fighting all the time！！

A. KitayutaMart2

水题～求一下公式即可，2^(n+1) = T/k+1.

CODE :

#line 7 "KitayutaMart2.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 KitayutaMart2    {            public:            int numBought(int K, int T)            {                    //$CARETPOSITION$                    int x = T/K+1;                    int n = 0, s = 1;                    while(s < x) {                        s *= 2;                        n++;                    }                    return n;            }};

B.Fragile2

由于数据量很小，枚举两个删除点，在dfs一下图，即可。

CODE；

#line 7 "Fragile2.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;bool vis[55];int gra[60][60], g[60][60];int n;class Fragile2{        public:        void init(VS graph)        {            for(int i = 0; i < n; ++i) {                for(int j = 0; j < n; ++j) {                    if(graph[i][j] == 'Y') gra[i][j] = 1;                    else gra[i][j] = 0;                }            }        }        int dfs(int st)        {            for(int i = 0; i < n; ++i) {                if(!vis[i] && g[st][i]) {                    vis[i] = 1;                    dfs(i);                }            }        }        int work(int x, int y)        {            memset(vis, 0, sizeof(vis));            for(int i = 0; i < n; ++i) {                for(int j = 0; j < n; ++j) {                    g[i][j] = gra[i][j];                }            }            vis[x] = 1; vis[y] = 1;            for(int i = 0; i < n; ++i) {                g[x][i] = 0; g[i][x] = 0;                g[y][i] = 0; g[i][y] = 0;            }            int cnt = 0;            for(int i = 0; i < n; ++i) {                if(!vis[i]) {                    vis[i] = 1;                    cnt++;                    dfs(i);                }            }            return cnt;        }        int countPairs(vector <string> graph)        {                int ans = 0;                n = graph.size();                memset(gra, 0, sizeof(gra));                memset(g, 0, sizeof(g));                init(graph);                int sum = work(n, n);                for(int i = 0; i < n; ++i) {                    for(int j = i + 1; j < n; ++j) {                        if(sum < work(i, j)) ans++;                    }                }                return ans;        }};

C.ABC

Way1：DP: dp[i][j][k][num] = 第i个位置j个A，k个B得到num对。

Way2：构造。枚举N个A，N个B, N个C, 三个for，直到（i×（j+k）+ j * k） == K 为止，可以连续输出A，B，C.

CODE :

#line 7 "ABC.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;short dp[32][35][35][500];class ABC{        public:        string cal(int x)        {            switch(x){                case 0: return "A";                case 1: return "B";                case 2: return "C";            }        }        string createString(int N, int K)        {            memset(dp, -1, sizeof(dp));            dp[0][0][0][0] = 0;            for(int i = 0; i < N; ++i) {                for(int j = 0; j <= i; ++j) {                    for(int k = 0; k <= i; ++k) {                        if(j + k > i) continue;                        for(int num = 0; num <= max(min((i*(i-1))/2, K), 0); ++num) {                            if(dp[i][j][k][num] < 0) continue;                            //printf("%d %d %d %d\n", i, j, k, num);                            for(int op = 0; op < 3; ++op) {                                if(op == 0)                                    dp[i+1][j+1][k][num+i-j] = op;                                else if(op == 1)                                    dp[i+1][j][k+1][num+i-(j+k)] = op;                                else                                    dp[i+1][j][k][num] = op;                            }                        }                    }                }            }            string ans;            int k = K;            for(int i = 0; i <= N; ++i) {                for(int j = 0; i+j <= N; ++j) {                    if(dp[N][i][j][k] >= 0) {                        while(N) {                            int d = dp[N][i][j][k];                            string op = cal(d);                            ans += op;                            --N;                            if(d == 0) {                                --i;                                k -= (N-i);                            }                            else if(d == 1) {                                --j;                                k -= (N-i-j);                            }                        }                        return ans;                    }                }            }            return ans;        }};

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