ZOJ 3209 Treasure Map

最新推荐文章于 2021-08-04 15:15:41 发布

wjf1514

最新推荐文章于 2021-08-04 15:15:41 发布

阅读量656

点赞数

CC 4.0 BY-SA版权

分类专栏： DLX

本文链接：https://blog.youkuaiyun.com/dotacm/article/details/12235509

DLX 专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一种基于DLX算法的解决方案，该算法能够高效地处理在一个大矩形区域内放置多个小矩形的问题，确保这些小矩形既不重叠也不遗漏区域的任何部分。通过详细的代码实现和解释，读者可以了解如何利用DLX算法来解决此类覆盖问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

DLX。用尽量少的矩形，覆盖一个大矩形，不重不漏，求使用个数。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<string>
#include<queue>
#include<map>
///LOOP
#define REP(i, n) for(int i = 0; i < n; i++)
#define FF(i, a, b) for(int i = a; i < b; i++)
#define FFF(i, a, b) for(int i = a; i <= b; i++)
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)
#define FDD(i, a, b) for(int i = a; i >= b; i--)
///INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RFI(n) scanf("%lf", &n)
#define RFII(n, m) scanf("%lf%lf", &n, &m)
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
#define RS(s) scanf("%s", s)
///OUTPUT
#define PN printf("\n")
#define PI(n) printf("%d\n", n)
#define PIS(n) printf("%d ", n)
#define PS(s) printf("%s\n", s)
#define PSS(s) printf("%s ", s)
#define PC(n) printf("Case %d: ", n)
///OTHER
#define PB(x) push_back(x)
#define CLR(a, b) memset(a, b, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(b))
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = 1e9+7;
const int INFI = 1e9 * 2;
const LL LINFI = 1e17;
const double eps = 1e-6;
const int N = 1111;
const int M = 111111;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};

int U[M], D[M], R[M], L[M], Y[M], H[N], S[N];
int size, ans, n, m;

void remove(int c)
{
    L[R[c]] = L[c];
    R[L[c]] = R[c];
    for(int i = D[c]; i != c; i = D[i])
        for(int j = R[i]; j != i; j = R[j])
            U[D[j]] = U[j], D[U[j]] = D[j], S[Y[j]]--;
}

void resume(int c)
{
    R[L[c]] = L[R[c]] = c;
    for(int i = U[c]; i != c; i = U[i])
        for(int j = L[i]; j != i; j = L[j])
            S[Y[U[D[j]] = D[U[j]] = j]]++;
}

void dance(int cnt)
{
    if(cnt >= ans)return ;
    if(!R[0])
    {
        ans = min(ans, cnt);
        return ;
    }
    int k, p;
    k = INFI;
    for(int i = R[0]; i; i = R[i])if(k > S[i])k = S[p = i];
    remove(p);
    for(int i = D[p]; i != p; i = D[i])
    {
        for(int j = R[i]; j != i; j = R[j])remove(Y[j]);
        dance(cnt + 1);
        for(int j = L[i]; j != i; j = L[j])resume(Y[j]);
    }
    resume(p);
    return ;
}

void link(int r, int c)
{
    S[Y[size] = c]++;
    D[size] = D[U[size] = c];
    D[c] = U[D[size]] = size;
    if(H[r] < 0)H[r] = L[size] = R[size] = size;
    else
    {
        R[size] = R[L[size] = H[r]];
        R[H[r]] = L[R[size]] = size;
    }
    size++;
}

void init(int k)
{
    REP(i, k)R[U[i] = D[i] = L[i + 1] = i] = i + 1;
    R[k - 1] = 0;
    size = k;
    CLR(H, -1);
    CLR(S, 0);
    ans = INFI;
}

int main()
{
    //freopen("input.txt", "r", stdin);

    int x1, x2, y1, y2, k, t;
    RI(t);
    while(t--)
    {
        RIII(n, m, k);
        init(n * m + 1);
        FFF(i, 1, k)
        {
            RIV(x1, y1, x2, y2);
            FF(p, x1, x2)FF(q, y1, y2)link(i, p * m + q + 1);
        }
        dance(0);
        PI(ans == INFI ? -1 : ans);
    }
    return 0;
}