CUGOJ 1437 Take Photo

题目

Description

There are n people standing in a line to take photo. But for some reasons ,there are some limits on their positions. For example, i must stand on j’s right, marked for (i, j). Now, giving you the n people and those limits, please tell me the number of the legal permutation.

Input

There are multiple test cases. For each test:
The first line contains one positive integer n (1<=n<=16), indicating the number of people. Then following there are n lines, each line containing n integers. The element at ith row and jth column is either 1 or 0; if the element is 1, it represents the limit (i, j); You can assume that any pair of (i, j) is bound to be 0 if i is equal to j.
Processed to the end of file.

Output

 For each test case, output the number of legal permutation on one line.

Sample Input

2

0 1

1 0

3

0 0 0

0 0 0

0 0 0

3

0 0 0

1 0 0

0 1 0

Sample Output

0

6

1

状态dp，类似于棋盘问题，加上一些限制，状态i中1的个数表示已经安排了几个人，并且放在了哪些行，

// http://acm.cug.edu.cn/JudgeOnline/problem.php?id=1437#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;typedef long long ll;ll dp[1<<16];int n;int map[17][17];int num1[17],num2[17],num3[17],num4[17];inline int getone(int i){    int cnt=0;    while(i) cnt+=i%2,i>>=1;    return cnt;}int main(){    while(~scanf("%d",&n))    {        memset(num1,0,sizeof(num1));        memset(num3,0,sizeof(num3));        memset(num4,0,sizeof(num4));        memset(num2,0,sizeof(num2));        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)            {                scanf("%d",&map[i][j]);            }        int tag=1;        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)                if(i!=j)                {                    if(map[i][j]&&map[j][i]) tag=0;break;                }        if(!tag)        {            puts("0");            continue;        }        for(int i=0;i<n;i++)        {            for(int j=0;j<i;j++)            {                if(map[j][i]) num1[i]++;//在0～i人中有多少个人必须在i的右边                if(map[i][j]) num2[i]++;//在0~i人中有多少个人必须在i的左边            }        }        for(int i=0;i<n;i++)        {            for(int j=i+1;j<n;j++)            {                if(map[j][i]) num3[i]++;//在i+1~n-1人中。。。。。                if(map[i][j]) num4[i]++;//。。。。            }        }        int all=(1<<n)-1;        memset(dp,0,sizeof(dp));        dp[0]=1;        for(int i=1;i<=all;i++)        {            int rr=getone(i)-1;            ll ans=0;            for(int j=0;j<n;j++)            {                if(i&(1<<j))                {                    int l=0,r=0,ll=0,rrr=0;                    for(int k=0;k<j;k++)                        (i&(1<<k))?l++:ll++;                    for(int k=j+1;k<n;k++)                        (i&(1<<k))?r++:rrr++;                    if(r<num1[rr]) continue;                    if(l<num2[rr]) continue;                    if(rrr<num3[rr]) continue;                    if(ll<num4[rr]) continue;                    ans+=dp[i&(~(1<<j))];                }            }            dp[i]=ans;        }        printf("%lld\n",dp[all]);    }    return 0;}