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-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 70;
const int maxNode = 5000;
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 = 0;
int vis[70];
mem(vis);
for(int i = R[0]; i!=0; i = R[i]) {
if(!vis[i]) {
cnt++;
vis[i] = 1;
for(int j = D[i]; j!=i; j = D[j]) {
for(int k = R[j]; k!=j; k = R[k]) {
vis[col[k]] = 1;
}
}
}
}
return cnt;
}
int dfs(int d) {
if(d+h()>k)
return 0;
if(R[0] == 0) {
return 1;
}
int c = R[0];
for(int i = R[0]; i!=0; 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+1))
return 1;
for(int j = L[i]; j!=i; j = L[j])
resume(j);
resume(i);
}
return 0;
}
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 = 0; i<=n; i++) {
R[i] = i+1;
L[i] = i-1;
U[i] = i;
D[i] = i;
}
mem(S);
R
= 0;
L[0] = n;
sz = n;
}
int check(ll x) {
init();
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
if(dis[i][j] <= x) {
add(i, j);
}
}
}
if(dfs(0))
return 1;
return 0;
}
void solve() {
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; i++) {
scanf("%d%d", &x[i], &y[i]);
}
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
dis[i][j] = 1LL*abs(x[i]-x[j])+abs(y[i]-y[j]);
}
}
ll l = 0, r = 4e9+7, ans;
for(int i = 0; i < 50; i++) {
ll mid = l+r>>1LL;
if(check(mid)) {
ans = mid;
r = mid-1;
} else {
l = mid+1;
}
}
printf("%lld\n", ans);
}
}dlx;
int main()
{
int t;
cin>>t;
for(int i = 1; i<=t; i++) {
printf("Case #%d: ", i);
dlx.solve();
}
return 0;
}