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

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
[i][j][k] >= 0) {
while(N) {
int d = dp
[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;
}
};