HDU 5046 Airport【DLX重复覆盖】

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题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5046

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，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，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！
hdu 2295 和这题一个套路，更裸一些。
跳舞链好强大，可惜只会用模板，这个讲的还挺清晰

代码：

/*************************************************************************    > File Name: O.cpp    > Author: jiangyuzhu    > Mail: 834138558@qq.com    > Created Time: Fri 08 Jul 2016 03:31:58 PM CST ************************************************************************/#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<map>using namespace std;#define sal(n) scanf("%I64d", &(n))#define sa(n) scanf("%d", &(n))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];int S[maxc], H[maxr], size;int n, m, k;ll d[maxr][maxc], t[maxn];///不需要S域void Link(int r, int c){    S[c]++; C[size] = c;    U[size] = U[c]; D[U[c]] = size;    D[size] = c; U[c] = size;    U[c] = size;    if(H[r] == -1) H[r] = L[size] = R[size] = size;    else {        L[size] = L[H[r]]; R[L[H[r]]] = size;        R[size] = H[r]; L[H[r]] = size;    }    size++;}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{                   if(a + h() > k) return 0;    if(!R[0]) return a <=  k;    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;        for (int j = L[i]; j != i; j = L[j])            resume(j);        resume(i);    }    return 0;}void initL(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;    size = 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}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){    initL(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);}int main (void){    int T;sa(T);    for(int tt = 1; tt <= T; tt++){        sa(n);sa(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;        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", tt, t[r]);    }    return 0;}

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