UVA 10160 Servicing Stations

UVA_10160

    这个题目我改下一下Dancing Links的算法，没想到搜过去了，只不过时间花的比较多。

    由于原本Dancing Links是解决精确覆盖问题的，而这个问题是可以重复覆盖的，所以要把Dancing Links的删除和恢复的操作改变一下。

    当然，还有更高效的搜索方式，可以google一下。

#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f3f
const int MAXN = 40;
const int MAXD = MAXN * MAXN + MAXN;
int U[MAXD], D[MAXD], L[MAXD], R[MAXD], C[MAXD], H[MAXN], S[MAXN];
int vis[MAXN], del[MAXN][MAXN], N, M, Min, size, g[MAXN][MAXN];
void prepare(int r, int c)
{
    int i;
    for(i = 0; i <= c; i ++)
    {
        U[i] = D[i] = i;
        L[i + 1] = i, R[i] = i + 1;
        S[i] = 0;
    }
    R[c] = 0;
    while(r)
        H[r --] = -1;
    size = c;
}
void insert(int r, int c)
{
    ++ size;
    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
    {
        L[size] = H[r];
        R[size] = R[H[r]];
        L[R[H[r]]] = size;
        R[H[r]] = size;
    }
}
void remove(int c)
{
    R[L[c]] = R[c];
    L[R[c]] = L[c];
}
void resume(int c)
{
    R[L[c]] = c;
    L[R[c]] = c;
}
void dance(int cur)
{
    int i, j, c, min = INF;
    if(!R[0])
    {
        if(cur < Min)
            Min = cur;
        return ;
    }
    if(cur >= Min - 1)
        return ;
    for(i = R[0]; i != 0; i = R[i])
        if(S[i] < min)
        {
            min = S[i];
            c = i;
        }
    del[cur][c] = vis[c] = 1;
    remove(c);
    for(i = D[c]; i != c; i = D[i])
    {
        for(j = R[i]; j != i; j = R[j])
            if(!vis[C[j]])
            {
                vis[C[j]] = del[cur][C[j]] = 1;
                remove(C[j]);
            }
        dance(cur + 1);
        for(j = L[i]; j != i; j = L[j])
            if(del[cur][C[j]])
            {
                resume(C[j]);
                vis[C[j]] = del[cur][C[j]] = 0;
            }
    }
    resume(c);
    del[cur][c] = vis[c] = 0;
}
void solve()
{
    int i, j, k, x, y;
    prepare(N, N);
    memset(g, 0, sizeof(g));
    for(i = 1; i <= N; i ++)
        g[i][i] = 1;
    for(i = 0; i < M; i ++)
    {
        scanf("%d%d", &x, &y);
        g[x][y] = g[y][x] = 1;
    }
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= N; j ++)
            if(g[i][j])
                insert(i, j);
    Min = INF;
    memset(del, 0, sizeof(del));
    memset(vis, 0, sizeof(vis));
    dance(0);
    printf("%d\n", Min);
}
int main()
{
    for(;;)
    {
        scanf("%d%d", &N, &M);
        if(!N && !M)
            break;
        solve();
    }
    return 0;
}