UVA 387 A Puzzling Problem

最新推荐文章于 2014-10-25 16:42:26 发布

转载最新推荐文章于 2014-10-25 16:42:26 发布 · 100 阅读

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原文链接：http://www.cnblogs.com/staginner/archive/2011/12/23/2298785.html

UVA_387

由于是精确覆盖问题，所以可以用Dancing Links去解决，只是需要把每个拼图按摆放位置的不同分成若干个拼图，但这些拼图都属于一类的，这一点要在Dancing Links的列中体现出来。

#include<stdio.h>
#include<string.h>
#define MAXM 6
#define INF 0x3f3f3f3f
const int MAXN = 6 * 6;
const int MAXD = 6 * 6 * 6 * (6 * 6 + 6) + 6 * 6 + 6;
char b[MAXM];
int N = 4, M, ans[MAXM][MAXM], g[MAXM][MAXM][MAXM], x[MAXM], y[MAXM];
int size, U[MAXD], D[MAXD], L[MAXD], R[MAXD], S[MAXN], H[MAXD], X[MAXD], C[MAXD], Q[MAXD];
int st[MAXD], sx[MAXD], sy[MAXD];
void prepare(int r, int c)
{
int i;
for(i = 0; i <= c; i ++)
    {
        L[i + 1] = i, R[i] = i + 1;
        S[i] = 0;
        U[i] = D[i] = i;
    }
    R[c] = 0;
    size = c;
while(r)
        H[r --] = -1;
}
void insert(int r, int c)
{
    ++ size;
    X[size] = r;
    C[size] = c;
    ++ S[c];
    U[size] = c;
    D[size] = D[c];
    U[D[c]] = size;
    D[c] = size;
if(H[r] == -1)
        H[r] = L[size] = R[size] = size;
else
    {
        R[size] = R[H[r]];
        L[size] = H[r];
        L[R[H[r]]] = size;
        R[H[r]] = size;
    }
}
void init()
{
int i, j, k, p, q, row;
    prepare(N * N * M, N * N + M);
    row = 0;
for(i = 1; i <= M; i ++)
    {
        scanf("%d%d", &x[i], &y[i]);
for(j = 0; j < x[i]; j ++)
        {
            scanf("%s", b);
for(k = 0; k < y[i]; k ++)
                g[i][j][k] = b[k] - '0';
        }
for(j = 0; j + x[i] <= N; j ++)
for(k = 0; k + y[i] <= N; k ++)
            {
                ++ row;
                st[row] = i, sx[row] = j, sy[row] = k;
                insert(row , i);
for(p = 0; p < x[i]; p ++)
for(q = 0; q < y[i]; q ++)
if(g[i][p][q])
                            insert(row, (j + p) * N + (k + q) + M + 1);
            }
    }
}
void remove(int c)
{
int i, j;
    L[R[c]] = L[c];
    R[L[c]] = R[c];
for(i = D[c]; i != c; i = D[i])
for(j = R[i]; j != i; j = R[j])
        {
            U[D[j]] = U[j];
            D[U[j]] = D[j];
        }
}
void resume(int c)
{
int i, j;
for(i = U[c]; i != c; i = U[i])
for(j = L[i]; j != i; j = L[j])
        {
            U[D[j]] = j;
            D[U[j]] = j;
        }
    L[R[c]] = c;
    R[L[c]] = c;
}
int dance(int cur)
{
int i, j, min, c;
if(!R[0])
    {
int k, row, p, q;
for(i = 0; i < cur; i ++)
        {
            row = Q[i];
            k = st[row];
for(p = 0; p < x[k]; p ++)
for(q = 0; q < y[k]; q ++)
if(g[k][p][q])
                        ans[p + sx[row]][q + sy[row]] = k;
        }
return 1;
    }
    min = INF;
for(i = R[0]; i != 0; i = R[i])
if(S[i] < min)
        {
            min = S[i];
            c = i;
        }
    remove(c);
for(i = D[c]; i != c; i = D[i])
    {
        Q[cur] = X[i];
for(j = R[i]; j != i; j = R[j])
            remove(C[j]);
if(dance(cur + 1))
return 1;
for(j = L[i]; j != i; j = L[j])
            resume(C[j]);
    }
    resume(c);
return 0;
}
void solve()
{
int i, j;
if(dance(0))
    {
for(i = 0; i < N; i ++)
        {
for(j = 0; j < N; j ++)
                printf("%d", ans[i][j]);
            printf("\n");
        }
    }
else
        printf("No solution possible\n");
}
int main()
{
int t = 0;
for(;;)
    {
        scanf("%d", &M);
if(!M)
break;
if(t ++)
            printf("\n");
        init();
        solve();
    }
return 0;
}

转载于:https://www.cnblogs.com/staginner/archive/2011/12/23/2298785.html