HDU 5046 Airport（二分+DLX重复覆盖）

题意：

给定n个城市的坐标，要在城市中建k个飞机场，使城市距离最近的飞机场的最长距离最小，求这个最小距离。

分析：

最小化最大值，显然二分最大距离。然后我们将距离在范围内的两个城市建边，建一个n*n的矩阵，看能否选出k个城市，使得覆盖了所有城市。 将点之间的关系转化成01矩阵的覆盖问题，每一行代表一个城市可以覆盖到的城市，重复覆盖，建好边套个DLX即可。 
看了鸟神博客，这里可以直接将所有距离保存在一个数组中，排序并去重，二分下标即可。这样快了很多很多！ 

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<map>using namespace std;typedef long long ll;const int maxn = 3600 + 5, maxc = 60 + 5, maxr = 60 + 5, inf = 0x3f3f3f3f;struct Node{ll x; ll y;}city[maxn];int L[maxn], R[maxn], D[maxn], U[maxn], C[maxn]; // 记录某个标号的节点的上下左右节点的编号int S[maxc], H[maxr], sz;  // H记录每行节点的信息(无行头)，S保存某一列1的数量int n, m, k;ll d[maxr][maxc], t[maxn];///不需要S域void init(int x)///col is 1~x,row start from 1{    for (int i = 0; i <= x; ++i)    {        S[i] = 0;        D[i] = U[i] = i;        L[i+1] = i; R[i] = i+1;    }///对列表头初始化    R[x] = 0;    sz = x + 1;///真正的元素从m+1开始    memset (H, -1, sizeof(H));    ///mark每个位置的名字}void Link(int r, int c){    S[c]++; C[sz] = c;    U[sz] = U[c]; D[U[c]] = sz;    D[sz] = c; U[c] = sz;    U[c] = sz;    if(H[r] == -1) H[r] = L[sz] = R[sz] = sz;    else    {        L[sz] = L[H[r]]; R[L[H[r]]] = sz;        R[sz] = H[r]; L[H[r]] = sz;    }    sz++;}void remove(int c){    for(int i = D[c]; i != c; i = D[i])        L[R[i]] = L[i], R[L[i]] = R[i];}void resume(int c){    for (int i = U[c]; i != c; i = U[i])        L[R[i]] = R[L[i]] = i;}int h(){///用精确覆盖去估算剪枝    int ret = 0;    bool vis[maxc];    memset (vis, false, sizeof(vis));    for(int i = R[0]; i; i = R[i])    {        if(vis[i]) continue;        ret++;        vis[i] = true;        for (int j = D[i]; j != i; j = D[j])            for (int k = R[j]; k != j; k = R[k])                vis[C[k]] = true;    }    return ret;}int ans;bool Dance(int a) //具体问题具体分析, a是递归层数{    if(a + h() > k) return 0;  //如果 当前层数+即将要加的层数 > 说明肯定不可以    if(!R[0]) return a <= k;  //说明出答案了    int c = R[0];    for (int i = R[0]; i; i = R[i])        if(S[i] < S[c]) c = i;    for(int i = D[c]; i != c; i = D[i])    {        remove(i);        for(int j = R[i]; j != i; j = R[j])            remove(j);        if(Dance(a + 1)) return 1;        // 这里必须反着恢复        for (int j = L[i]; j != i; j = L[j])            resume(j);        resume(i);    }    return 0;}ll dist(int i, int j){    ll d = abs(city[i].x - city[j].x) +  abs(city[i].y - city[j].y);    return d;}bool judge(ll mid){    init(n);    ans = 0x3f3f3f3f;    for(int i = 1; i <= n; i++)    {        for(int j = 1; j <= n; j++)            if(d[i][j] <= mid)                Link(i, j);    }    return Dance(0);  //从0层开始}int main (void){    int T, ca = 1;    cin >> T;    while(T--)    {        scanf("%d%d", &n, &k);        ll maxd = 0;        int cnt = 0;        for(int i = 1; i <= n; i++)            scanf("%I64d%I64d", &city[i].x, &city[i].y);        for(int i = 1; i <= n; i++)            for(int j = 1; j <= n; j++)            {                d[i][j] = dist(i, j);                t[cnt++] = d[i][j];            }        sort(t, t + cnt);        int ncnt = unique(t, t + cnt) - t;  //去重， 二分所有可能的距离更好        ll l = -1, r = ncnt;        while(r - l > 1)        {            ll mid = (l + r) / 2;            if(judge(t[mid])) r = mid;            else l = mid;        }        printf("Case #%d: %I64d\n", ca++, t[r]);    }    return 0;}