hdu 5046 Airport 二分+重复覆盖

题目链接

给n个点， 定义两点之间距离为|x1-x2|+|y1-y2|。 然后要选出k个城市建机场， 每个机场可以覆盖一个半径的距离。 求在选出点数不大于k的情况下， 这个半径距离的最大值。

二分半径， 然后距离小于等于半径的就连边， 然后跑重复覆盖。

#include<bits/stdc++.h>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, a, n) for(int i = a; i<n; i++)
#define ull unsigned long long
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-;
const int mod = 1e9+;
const int inf = ;
const int dir[][] = { {-, }, {, }, {, -}, {, } };
const int maxn = ;
const int maxNode = ;
int num;
struct DLX {
    int L[maxNode], R[maxNode], U[maxNode], D[maxNode], row[maxNode], col[maxNode];
    int S[maxn], H[maxn], sz, n, m, k, x[maxn], y[maxn];
    ll dis[maxn][maxn];
    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]] = i;
            R[L[i]] = i;
        }
    }
    int h() {
        int cnt = ;
        int vis[];
        mem(vis);
        for(int i = R[]; i!=; i = R[i]) {
            if(!vis[i]) {
                cnt++;
                vis[i] = ;
                for(int j = D[i]; j!=i; j = D[j]) {
                    for(int k = R[j]; k!=j; k = R[k]) {
                        vis[col[k]] = ;
                    }
                }
            }
        }
        return cnt;
    }
    int dfs(int d) {
        if(d+h()>k)
            return ;
        if(R[] == ) {
            return ;
        }
        int c = R[];
        for(int i = R[]; i!=; i = R[i])
            if(S[c]>S[i])
                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(dfs(d+))
                return ;
            for(int j = L[i]; j!=i; j = L[j])
                resume(j);
            resume(i);
        }
        return ;
    }
    void add(int r, int c) {
        sz++;
        row[sz] = r;
        col[sz] = c;
        S[c]++;
        U[sz] = U[c];
        D[sz] = c;
        D[U[c]] = sz;
        U[c] = sz;
        if(~H[r]) {
            R[sz] = H[r];
            L[sz] = L[H[r]];
            L[R[sz]] = sz;
            R[L[sz]] = sz;
        } else {
            H[r] = L[sz] = R[sz] = sz;
        }
    }
    void init() {
        mem1(H);
        for(int i = ; i<=n; i++) {
            R[i] = i+;
            L[i] = i-;
            U[i] = i;
            D[i] = i;
        }
        mem(S);
        R[n] = ;
        L[] = n;
        sz = n;
    }
    int check(ll x) {
        init();
        for(int i = ; i <= n; i++) {
            for(int j = ; j <= n; j++) {
                if(dis[i][j] <= x) {
                    add(i, j);
                }
            }
        }
        if(dfs())
            return ;
        return ;
    }
    void solve() {
        scanf("%d%d", &n, &k);
        for(int i = ; i <= n; i++) {
            scanf("%d%d", &x[i], &y[i]);
        }
        for(int i = ; i <= n; i++) {
            for(int j = ; j <= n; j++) {
                dis[i][j] = 1LL*abs(x[i]-x[j])+abs(y[i]-y[j]);
            }
        }
        ll l = , r = 4e9+, ans;
        for(int i = ; i < ; i++) {
            ll mid = l+r>>1LL;
            if(check(mid)) {
                ans = mid;
                r = mid-;
            } else {
                l = mid+;
            }
        }
        printf("%lld\n", ans);
    }
}dlx;
int main()
{
    int t;
    cin>>t;
    for(int i = ; i<=t; i++) {
        printf("Case #%d: ", i);
        dlx.solve();
    }
    return ;
}