Topcoder SRM 638 DIV 2 (Strength can produce miracles)_html/css_WEB-ITnose

The water question is a violence. Power works miracles.

Time limit (s):2.000Memory limit (MB):256Constraints sizeEach element in size-maxSizeSumExamples 1)

Problem Statement

There is a narrow passage. Inside the passage there are some wolves. You are given a vector size that contains the sizes of those wolves, from left to right.


The passage is so narrow that some pairs of wolves cannot pass by each other. More precisely, two adjacent wolves may swap places if and only if the sum of their sizes is maxSizeSum or less. Assuming that no wolves leave the passage, what is the number of different permutations of wolves in the passage? Note that two wolves are considered different even if they have the same size.


Compute and return the number of permutations of wolves that can be obtained from their initial order by swapping a pair of wolves zero or more times.

Definition

Class:	NarrowPassage2Easy
Method:	count
Parameters:	vector , int
Returns:	int
Method signature:	int count(vector size, int maxSizeSum)
(be sure your method is public)

Limits

Time limit (s):	2.000
Memory limit (MB):	256

-

will contain between 1 and 6 elements, inclusive.

-

will be between 1 and 1,000, inclusive.

will be between 1 and 1,000, inclusive.

0)

{1, 2, 3}

Returns: 2

From {1, 2, 3}, you can swap 1 and 2 to get {2, 1, 3}. But you can't get other permutations.

{1, 2, 3}

1000

Returns: 6

From {1, 2, 3}, you can swap 1 and 2 to get {2, 1, 3}. But you can't get other permutations.

Here you can swap any two adjacent wolves. Thus, all 3! = 6 permutations are possible.

{1, 2, 3}

Returns: 3

You can get {1, 2, 3}, {2, 1, 3} and {2, 3, 1}.

Here you can swap any two adjacent wolves. Thus, all 3! = 6 permutations are possible.

	2) {1,1,1,1,1,1} Returns: 720 All of these wolves are different, even though their sizes are the same. Thus, there are 6! different permutations possible.

{2,4,6,1,3,5}

You can get {1, 2 , 3}, {2, 1, 3} and {2, 3, 1}. 3) All of these wolves are different, even though their sizes are the same. Thus, there are 6! different permutations possible. 4)

Returns: 60

5)

{1000}

1000

Returns: 1

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define eps 1e-10///#define M 1000100///#define LL __int64#define LL long long#define INF 0x7fffffff#define PI 3.1415926535898#define zero(x) ((fabs(x)<eps)?0:x)using namespace std;const int maxn = 1000010;int vis[10][10];class NarrowPassage2Easy{public:    int count(vector <int> size, int maxSizeSum)    {        int len = size.size();        memset(vis, 0, sizeof(vis));        for(int i = 1; i <= len; i++)        {            for(int j = i+1; j <= len; j++)            {                int x = size[i-1]+size[j-1];                if(x > maxSizeSum)                {                    ///vis[i][j] = 1;                    vis[j][i] = 1;                }            }        }        for(int i = 1; i <= len; i++)        {            for(int j = 1; j <= len; j++)            {                cout<<vis[i][j]<<" ";            }            cout<<endl;        }        int sum = 0;        if(len == 1) return 1;        if(len == 2)        {            if(size[0]+size[1] > maxSizeSum) return 1;            return 2;        }        if(len == 3)        {            for(int i = 1; i <= len; i++)            {                for(int j = 1; j <= len; j++)                {                    if(i == j) continue;                    for(int k = 1; k <= len; k++)                    {                        if(k == i || k == j) continue;                        if(vis[i][j] || vis[i][k] || vis[j][k]) continue;                        sum++;                    }                }            }        }        if(len == 4)        {            for(int i1 = 1; i1 <= len; i1++)            {                for(int i2= 1; i2 <= len; i2++)                {                    if(i1 == i2) continue;                    for(int i3 = 1; i3 <= len; i3++)                    {                        if(i1 == i3 || i2 == i3) continue;                        for(int i4 = 1; i4 <= len; i4++)                        {                            if(i1 == i4 || i2 == i4 || i3 == i4) continue;                            if(vis[i1][i2] || vis[i1][i3] || vis[i1][i4]                                       || vis[i2][i3] ||vis[i2][i4]                                       ||vis[i3][i4]) continue;                                sum++;                        }                    }                }            }        }        if(len == 5)        {            for(int i1 = 1; i1 <= len; i1++)            {                for(int i2= 1; i2 <= len; i2++)                {                    if(i1 == i2) continue;                    for(int i3 = 1; i3 <= len; i3++)                    {                        if(i1 == i3 || i2 == i3) continue;                        for(int i4 = 1; i4 <= len; i4++)                        {                            if(i1 == i4 || i2 == i4 || i3 == i4) continue;                            for(int i5 = 1; i5 <= len; i5++)                            {                                if(i1 == i5 || i2 == i5 || i3 == i5 || i4 == i5) continue;                                if(vis[i1][i2] || vis[i1][i3] || vis[i1][i4] || vis[i1][i5]                                       || vis[i2][i3] ||vis[i2][i4] || vis[i2][i5]                                       ||vis[i3][i4] || vis[i3][i5]                                       ||vis[i4][i5]) continue;                                sum++;                            }                        }                    }                }            }        }        if(len == 6)        {            for(int i1 = 1; i1 <= len; i1++)            {                for(int i2= 1; i2 <= len; i2++)                {                    if(i1 == i2) continue;                    for(int i3 = 1; i3 <= len; i3++)                    {                        if(i1 == i3 || i2 == i3) continue;                        for(int i4 = 1; i4 <= len; i4++)                        {                            if(i1 == i4 || i2 == i4 || i3 == i4) continue;                            for(int i5 = 1; i5 <= len; i5++)                            {                                if(i1 == i5 || i2 == i5 || i3 == i5 || i4 == i5) continue;                                for(int i6 = 1; i6 <= len; i6++)                                {                                    if(i1 == i6 || i2 == i6 || i3 == i6 || i4 == i6 || i6 == i5) continue;                                    if(vis[i1][i2] || vis[i1][i3] || vis[i1][i4] || vis[i1][i5] || vis[i1][i6]                                       || vis[i2][i3] ||vis[i2][i4] || vis[i2][i5] || vis[i2][i6]                                       ||vis[i3][i4] || vis[i3][i5] || vis[i3][i6]                                       ||vis[i4][i5] || vis[i4][i6]                                       || vis[i5][i6]) continue;                                   sum ++;                                }                            }                        }                    }                }            }        }        return sum;    }};int main(){    NarrowPassage2Easy a;    vector<int> f;    f.push_back(189);    f.push_back(266);    cout<<a.count(f, 186)<<endl;;    return 0;}