poj_1321

来源：互联网发布：淘宝v任务编辑：程序博客网时间：2024/05/22 12:04

源链接：http://poj.org/problem?id=1321

题目大意就是在n*n的棋盘上防止棋子，注意是'#'的地方才能放，且每行每列最多一个。

跟八皇后问题有点相似，我们只要逐行开始放，有一点注意，就是当k<n时我们可以选择某一行不放，所以在搜索的时候多一句话即可。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m,k;#define Mod 1000000007#define N 110#define M 1000100char mp[N][N];int vis[N];int ans;void dfs(int row,int num){if(num == 0){ans++;return;}if(row>=n)//防止搜索越界return;for(int i=0;i<n;i++){if(mp[row][i] == '#'&&!vis[i]){vis[i] = 1;dfs(row+1,num-1);vis[i] = 0;}}dfs(row+1,num);//row这行不放return;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endif//    int t;    while(sfd(n,k)!=EOF){    if(n==-1 && k==-1)    break;    for(int i=0;i<n;i++)    sfs(mp[i]);    int row[10];    int col[10];    memset(row,0,sizeof row);    memset(col,0,sizeof col);    for(int i=0;i<n;i++)    for(int j=0;j<n;j++)    if(mp[i][j] == '#'){    row[i] = 1;    col[j] = 1;    }    int res1 = 0;    int res2 = 0;    for(int i=0;i<n;i++){    if(row[i])    res1++;    if(col[i])    res2++;    }    if(res1<k || res2<k){    printf("0\n");    }else{    ans = 0;    memset(vis,0,sizeof vis);    dfs(0,k);    printf("%d\n",ans);    }    }return 0;}

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