2719: [Violet 4]银河之星

2719: [Violet 4]银河之星

Time Limit: 5 Sec   Memory Limit: 128 MB
Submit: 59   Solved: 35
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Description

Input

Output

Sample Input

Sample Output

HINT

Source

Pku 3490 

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显然？！。。棋盘上的棋子可以按照横纵坐标mod 3的余数分为9种。。然后跳两格的操作等价于拿走两种棋子然后放入一种新棋子。所以记忆化搜索暴力上个map。。然而棋盘不一定允许这种操作，所以预处理可行的交换。。(一开始只枚举给出的10个棋子，然后出错了。。。)然后判断类型的时候3*(x%3)一定要加括号。。。

总的可能性有C(18,8) < 50000 种

然后对于这个计算。。

等价于有9个篮子要往里面放10个苹果，允许篮子放空，但苹果一定要放完的方案数

先考虑没有篮子放空的情况：

10个苹果有9个空隙，只要放8个隔板就能分出9类C(9,8)

然后等价于

x1 + x2 + x3 + ... + x9 = 10

xi > 0

对于原问题

(x1+1) + (x2+1) + (x3+1) + ... + (x9+1) = 19

xi+1>0

就是18个空隙放8个隔板 C(18,8)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
using namespace std;
 
typedef long long LL;
const int dx[8] = {0,1,0,-1,1,-1,1,-1};
const int dy[8] = {1,0,-1,0,1,-1,-1,1};
const int ch[9][9] = {{0,2,1,6,8,7,3,5,4},{2,1,0,8,7,6,5,4,3},{1,0,2,7,6,8,4,3,5},
                      {6,8,7,3,5,4,0,2,1},{8,7,6,5,4,3,2,1,0},{7,6,8,4,3,5,1,0,2},
                      {3,5,4,0,2,1,6,8,7},{5,4,3,2,1,0,8,7,6},{4,3,5,1,0,2,7,6,8}};
 
map <LL,int> ma;
 
int n,m,k,X,Y,p[10][10],x[10],y[10];
LL now,goal,fac[10],che[10];
 
int ABS(int x) {return x < 0?-x:x;}
 
void Add(int X1,int Y1,int X2,int Y2)
{
	int A = 3*(X1%3) + Y1%3;
	int B = 3*(X2%3) + Y2%3;
	p[A][B] = p[B][A] = 1;
}
 
void Work(int a,int b)
{
	for (int l = 0; l < 8; l++) {
		int xa = a + dx[l]*3;
		int yb = b + dy[l]*3;
        if (xa < 1 || xa > n || yb < 1 || yb > m) continue;
		for (int i = 0; i < 8; i++) {
			int xx = xa + dx[i];
			int yy = yb + dy[i];
        	if (xx < 1 || xx > n || yy < 1 || yy > m) continue;
        	Add(a,b,xx,yy);
		}
	}
	for (int i = 0; i < 8; i++) {
		int xx = a + dx[i];
		int yy = b + dy[i];
       	if (xx < 1 || xx > n || yy < 1 || yy > m) continue;
       	Add(a,b,xx,yy);
	}
}
 
bool dfs(LL now,int tot)
{
    if (now == goal) return 1;
    if (ma[now]) return 0;
    else ma[now] = 1;
    if (tot == 1) return 0;
    for (int i = 0; i < 9; i++)
        for (int j = i + 1; j < 9; j++)
            if (p[i][j] && che[i] && che[j]) {
                --che[i]; --che[j]; ++che[ch[i][j]];
                LL nex = 0;
                for (int l = 0; l < 9; l++) nex += fac[l]*che[l];
                if (dfs(nex,tot-1)) return 1;
                ++che[i]; ++che[j]; --che[ch[i][j]];
            }
    return 0;
}
 
int main()
{
    #ifdef YZY
           freopen("galaxy4.in","r",stdin);
    #endif
     
    fac[0] = 1;
    for (int i = 1; i < 10; i++) fac[i] = fac[i-1]*11LL;
    while (scanf("%d%d%d%d%d",&k,&n,&m,&X,&Y) != EOF) {
        memset(p,0,sizeof(p)); memset(che,0,sizeof(che)); ma.clear();
        goal = fac[3*(X%3) + Y%3];
        for (int i = 0; i < k; i++) {
            scanf("%d%d",&x[i],&y[i]);
            ++che[3*(x[i]%3) + y[i]%3];
        }
        for (int i = 1; i <= n; i++)
        	for (int j = 1; j <= m; j++)
        		Work(i,j);
        LL st = 0;
        for (int i = 0; i < 9; i++) st += che[i]*fac[i];
        if (dfs(st,k)) printf("Yes\n");
        else printf("No\n");
    }
     
    return 0;
}