换topcoder

10.29

最近刷leetcode刷的有点膨胀了

点开web Arena做第一题AB，瞬间AC，卧槽我宝刀未老！

点开一个550、卧槽。。

点开一个850、卧槽草草草。。

弄好了Arena 准备起飞

850的题目是 问N个节点（各不相同）的图，有多少有座桥。

今天做了一题，POJ1737

我又用了一种乱七八糟的方法，我也是醉了

#include<iostream>
#include<algorithm>
#include<map>
#include<unordered_map>
#include<vector>
#include<string>
#include<stack>
#include<queue>

using namespace std;

const int MaxN = 50;

int main()
{
vector<long long> f = vector<long long>(MaxN + 1, 0);
vector<vector<long long>> g = vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0));
vector<vector<long long>> c = vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0));

c[0][0] = 1;
for (int i = 1; i <= MaxN; i++)
{
c[i][0] = c[i][i] = 1;
for (int j = 1; j <= i-1; j++)
c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
}

//
g[0][0] = 1;
//

f[1] = 1;
for (int n = 2; n <= MaxN; n++)
{
//g[n-1,n-1]
for (int k = 1; k <= n - 1; k++)
{
for (int s1 = 1; s1 <= n - 1; s1++)
g[n - 1][k] += ((1 << s1) - 1)*f[s1] * c[n - 2][s1 - 1] * g[n - 1 - s1][k - 1];
f
+= g[n - 1][k];
}
}

int X;
while (cin >> X)
{
if (!X) break;
cout << f[X] << endl;
}

return 0;
}

这当然是过不了了。。时间复杂度比较高，然后高精度也没有上。

不过意思到了就行了

我想的是 1连的那个连通分支有哪些点，用dp套一个dp

看了别人的解答：

设f(n)为所求答案

g(n)为n个顶点的非联通图

则f(n) + g(n) = h(n) = 2^(n * (n - 1) / 2)

其中h(n)是n个顶点的联图的个数

这样计算

先考虑1所在的连通分量包含哪些顶点

假设该连通分量有k个顶点

就有C(n - 1, k - 1)种集合

确定点集后，所在的连通分量有f(k)种情况。其他连通分量有 h(n - k)种情况

因此有递推公式。g(n) = sum{ C(n - 1, k - 1) * f(k) * h(n - k)} 其中k = 1,2...n-1

注意每次计算出g(n)后立刻算出f(n)

好吧 我那个做法也许是有一些借鉴意义的。。毕竟。。难想到。。

11.8

先贴一个小数据版本的，妈蛋为了搞这个一晚上没有玩多塔

#define _CRT_SECURE_NO_WARNINGS

#include<iostream>
#include<algorithm>
#include<map>
#include<unordered_map>
#include<vector>
#include<string>
#include<stack>
#include<queue>

using namespace std;

const int MaxN = 15;
const long long Modulo = 1000000007;

vector<vector<long long>> calcComb()
{
vector<vector<long long>> c = vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0));
c[0][0] = 1;
for (int i = 1; i <= MaxN; i++)
{
c[i][0] = c[i][i] = 1;
for (int j = 1; j <= i - 1; j++)
c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % Modulo;
}
return c;
}

long long pow(int s1, int c)
{
long long ret = 1;
while (c--) ret *= s1;
return ret;
}

int main()
{

auto C = calcComb();

//calc f1
: n nodes with 1 component
auto f1 = vector<long long>(MaxN + 1, 0);
auto g = vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0));
g[0][0] = 1;
f1[1] = 1;
for (int n = 2; n <= MaxN; n++)
{
for (int c = 1; c <= n - 1; c++)  // c components
{
for (int s1 = 1; s1 <= n - 1; s1++)
g[n - 1][c] += ((1 << s1) - 1) * f1[s1] * C[n - 2][s1 - 1] * g[n - 1 - s1][c - 1];
f1
+= g[n - 1][c];
}
}
//f2
: n nodes with no bridge
auto f2 = vector<long long>(MaxN + 1, 0);
g = vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0));
g[0][0] = 1;
f2[0] = 1;
for (int n = 1; n <= MaxN; n++)
{
f2
= f1
;
for (int c = 1; c <=  n - 1; c++)
{
for (int s1 = 1; s1 <= n - 1; s1++)
g[n - 1][c] += (s1 * C[n - 2][s1 - 1] * f1[s1] * g[n - 1 - s1][c - 1]);
}
for (int s1 = 1; s1 <= n - 1; s1++)
{
for (int c = 1; c <= n - 1;c++)
f2
-= (C[n - 1][s1 - 1] * f2[s1] * /*/ 激动的快哭了，找到个bug的时候，没有乘 /*/pow(s1,c) * g[n - s1][c]);
}
}

auto f3 = vector<vector<long long>>(MaxN + 1,vector<long long>(MaxN + 1,0));
//f3
[k] n nodes k bridges
f3[0][0] = 1;

auto gg = vector<vector<vector<long long>>>(MaxN + 1, vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0)));

gg[0][0][0] = 1;

for (int n = 1; n <= MaxN; n++)
{
f3
[0] = f2
;
for (int k = 1; k <= MaxN; k++)
{
for (int c = 1; c <= n - 1;c++)
for (int n1 = 1; n1 <= n - 1;n1++)
for (int k1 = 0; k1 < k; k1++)
{
gg[n - 1][k][c] += n1 * C[n-2][n1-1] * f3[n1][k1] * gg[n - 1 - n1][k - k1 - 1][c - 1];
}
for (int n1 = 1; n1 <= n - 1; n1++)
{
for (int c = 1; c <= n; c++)
f3
[k] += C[n - 1][n1 - 1] * f2[n1] * /*还TM漏一个*/pow(n1,c) * gg[n - n1][k][c];
}
}
}

auto f4 = vector<vector<vector<long long>>>(MaxN + 1, vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0)));
f4[0][0][0] = 1;
for (int n = 1; n <= MaxN; n++)
for (int k = 0; k < MaxN; k++)
{
f4
[k][1] = f3
[k];
for (int c = 2; c <= MaxN; c++)
{
for (int n1 = 1; n1 <= n;n1++)
for (int k1 = 0; k1 <= k; k1++)
f4
[k][c] += f3[n1][k1] * C[n - 1][n1 - 1] * f4[n - n1][k - k1][c-1];
}
}

int n, k;

while (cin >> n >> k)
{
int ret = 0;
for (int c = 1; c <= n; c++)
ret += f4
[k][c];
cout << ret << endl;
}

system("pause");
return 0;
}

再来一个超时版本的O(N^5)

#define _CRT_SECURE_NO_WARNINGS

#include<iostream>
#include<algorithm>
#include<map>
#include<unordered_map>
#include<vector>
#include<string>
#include<stack>
#include<queue>

using namespace std;

const int MaxN = 50;
const long long M = 1000000007;

long long mul(long long A, long long B)
{
return (A % M) * (B % M) % M;
}

void adt(long long& x, long long a)
{
x = (x + a) % M;
}

vector<vector<long long>> calcComb()
{
auto c = vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0));
c[0][0] = 1;
for (int i = 1; i <= MaxN; i++)
{
c[i][0] = c[i][i] = 1;
for (int j = 1; j <= i - 1; j++)
c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % M;
}
return c;
}

vector<vector<long long>> calcPow()
{
auto p = vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0));
for (int i = 1; i <= MaxN; i++)
{
p[i][0] = 1;
for (int j = 1; j <= MaxN; j++) p[i][j] = mul(i,p[i][j-1]);
}
return p;
}

int main()
{

auto C = calcComb();
auto POW = calcPow();

//calc f1
: n nodes with 1 component
auto f1 = vector<long long>(MaxN + 1, 0);
auto g = vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0));
g[0][0] = 1;
f1[1] = 1;
for (int n = 2; n <= MaxN; n++)
{
for (int c = 1; c <= n - 1; c++)  // c components
{
for (int s1 = 1; s1 <= n - 1; s1++)
g[n - 1][c] += mul(mul(mul((pow(2,s1) - 1),f1[s1]),C[n - 2][s1 - 1]),g[n - 1 - s1][c - 1]);
adt(f1
,g[n - 1][c]);
}
}
//f2
: n nodes with no bridge
auto f2 = vector<long long>(MaxN + 1, 0);
g = vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0));
g[0][0] = 1;
f2[0] = 1;
for (int n = 1; n <= MaxN; n++)
{
f2
= f1
;
for (int c = 1; c <=  n - 1; c++)
{
for (int s1 = 1; s1 <= n - 1; s1++)
adt(g[n - 1][c],mul(mul(mul(s1, C[n - 2][s1 - 1]),f1[s1]),g[n - 1 - s1][c - 1]));
}
for (int s1 = 1; s1 <= n - 1; s1++)
{
for (int c = 1; c <= n - 1;c++)
adt(f2
,M-mul(mul(mul(C[n - 1][s1 - 1],f2[s1]),/*/ 激动的快哭了，找到个bug的时候，没有乘 /*/POW[s1][c]),g[n - s1][c]));
}
}

auto f3 = vector<vector<long long>>(MaxN + 1,vector<long long>(MaxN + 1,0));
//f3
[k] n nodes k bridges
f3[0][0] = 1;

auto gg = vector<vector<vector<long long>>>(MaxN + 1, vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0)));

gg[0][0][0] = 1;

for (int n = 1; n <= MaxN; n++)
{
f3
[0] = f2
;
for (int k = 1; k <= MaxN; k++)
{
for (int c = 1; c <= n - 1;c++)
for (int n1 = 1; n1 <= n - 1;n1++)
for (int k1 = 0; k1 < k; k1++)
{
adt(gg[n - 1][k][c],mul(mul(mul(n1,C[n-2][n1-1]),f3[n1][k1]),gg[n - 1 - n1][k - k1 - 1][c - 1]));
}
for (int n1 = 1; n1 <= n - 1; n1++)
{
for (int c = 1; c <= n; c++)
adt(f3
[k],mul(mul(mul(C[n - 1][n1 - 1],f2[n1]), /*还TM漏一个*/POW[n1][c]),gg[n - n1][k][c]));
}
}
}

auto f4 = vector<vector<vector<long long>>>(MaxN + 1, vector<vector<long long>>(MaxN + 1, vector<long long>(MaxN + 1, 0)));
f4[0][0][0] = 1;
for (int n = 1; n <= MaxN; n++)
for (int k = 0; k < MaxN; k++)
{
f4
[k][1] = f3
[k];
for (int c = 2; c <= MaxN; c++)
{
for (int n1 = 1; n1 <= n;n1++)
for (int k1 = 0; k1 <= k; k1++)
adt(f4
[k][c],mul(mul(f3[n1][k1],C[n - 1][n1 - 1]),f4[n - n1][k - k1][c-1]));
}
}

int n, k;

while (cin >> n >> k)
{
long long ret = 0;
for (int c = 1; c <= n; c++)
adt(ret,f4
[k][c]);
cout << ret << endl;
}

system("pause");
return 0;
}