SCU2016-03 O题二分 + DLX可重复覆盖

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Analyse:
二分距离加DLX可重复覆盖。
注意这里的距离爆int.
Get:
在头文件
/**********************jibancanyang************************** *Author*        :jibancanyang *Created Time*  : 五  7/ 8 14:12:54 2016**Problem**:**Code**:***********************1599664856@qq.com**********************/#include <cstdio>#include <cstring>#include <iostream>#include <limits.h>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;typedef vector<int> vi;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a))#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3f3f3f3f;int n;int k; //最多选多少行const int maxnode = 10000;const int maxm = 65;const int maxn = 65;struct DLX{    int n,m,len;    int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];    int H[maxn];//行头结点    int S[maxm];//每列有多少个结点    int ansd,ans[maxn];//如果有答案，则选了ansd行，具体是哪几行放在ans[]数组里面,ans[0~ansd-1]    void init(int _n,int _m){        n = _n;m = _m;        for(int i = 0; i <= m; i++){            S[i] = 0;            U[i] = D[i] = i;//初始状态时，上下都指向自己            L[i] = i-1;            R[i] = i+1;        }        R[m] = 0,L[0] = m;        len = m;//编号，每列都有一个头结点，编号1~m        for(int i = 1; i <= n; i++)            H[i] = -1;//每一行的头结点    }    void link(int r,int c){//第r行，第c列        ++S[Col[++len]=c];//第len个节点所在的列为c,当前列的结点数++        Row[len] = r;//第len个结点行位置为r        D[len] = D[c];        U[D[c]] = len;        U[len] = c;        D[c] = len;        if(H[r] < 0)            H[r] = L[len] = R[len] = len;        else{            R[len] = R[H[r]];            L[R[H[r]]] = len;            L[len] = H[r];            R[H[r]] = len;        }    }    void del(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;    }    bool v[maxnode];    int f(){        int ret = 0;        for(int c = R[0]; c != 0; c = R[c])            v[c] = true;        for(int c = R[0]; c != 0; c = R[c]){            if(v[c]){                ret++;                v[c] = false;                for(int i = D[c]; i != c; i = D[i]){                    for(int j = R[i]; j != i; j = R[j]){                        v[Col[j]] = false;                    }                }            }        }        return ret;    }    bool dance(int d){//递归深度        if(d + f() > k)            return false;        if(R[0] == 0)            return d <= k;        int c = R[0];        for(int i = R[0]; i != 0; i = R[i]){            if(S[i] < S[c])                c = i;        }        for(int i = D[c]; i != c; i = D[i]){            del(i);            ans[d] = Row[i];//列头节点下面的一个节点            for(int j = R[i]; j != i; j = R[j])                del(j);            if(dance(d+1))                return true;            for(int j = L[i]; j != i; j = L[j])                resume(j);            resume(i);        }        return false;    }}head;/*int main(void) {    int n, m;  //n行m列的01矩阵中选取一些行，让每一列都至少含有一个1，最多选k行。    head.init(n, m);    if (matrix[i][j]矩阵单元值为1) head.link(i, j)    head.dance(0)；//返回是否可以覆盖的布尔值。}*/pii city[maxn];ll dist(pii &a, pii &b) {    return abs((ll)a.xx - b.xx) + abs((ll)a.yy - b.yy);}int main(void){#ifdef LOCAL    freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    //cout << INT_MAX << endl;    int T; sa(T);    for (int cas = 1; cas <= T; cas++) {        sa(n), sa(k);        for (int i = 0; i < n; i++) sa(city[i].xx), sa(city[i].yy);        ll l = 0, r = (ll)4e9 + 100;        while (l < r) {            ll mid = (r + l) / 2;             //        pr(l), pr(r), pl(mid);            head.init(n, n);            for (int i = 0; i < n; i++) {                for (int j = 0; j < n; j++) {                    if (dist(city[i], city[j]) <= mid)                         head.link(i + 1, j + 1);                }            }            if (head.dance(0)) r = mid;            else l = mid + 1;        }     //   printf("Case #%d: %lld\n", cas, l);        printf("Case #%d: %I64d\n", cas, l);    }    return 0;}
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