HDU 3656 Fire station（巧妙的二分+DLX重复覆盖）

题目地址
题意：给你n个城市，在这个n个城市中要修m个消防局，问在这个限制条件下使得每一个城市发生火灾所需要的最大救援长度最短，最短为多少。
思路：和HDU 2295 Radar一样，但是我们这里就不能直接二分枚举距离了，我看别人博客学到了一个巧妙的二分方法，因为要求最短，那么最短肯定是恰好是某个城市到消防局的距离，所有我们把城市与城市之间的距离离散出来，这样的话我们二分搜索下标，这样的话二分的时间就缩短了。

#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <cmath>#include <cstdio>#include <algorithm>#include <iomanip>#define N 1010#define MAXSIZE 1010*1010#define LL __int64#define inf 0x3f3f3f3f#define lson l,mid,ans<<1#define rson mid+1,r,ans<<1|1#define getMid (l+r)>>1#define movel ans<<1#define mover ans<<1|1using namespace std;const LL mod = 1000000007;const double eps = 1e-8;struct node {    int left, right, up, down, col, row;}mapp[MAXSIZE];int S[N], H[N];//S记录该列中1元素的个数int head, cnt;int len;int n, m;double lens[N][N], diss[N*N];//坑呀，居然忘记N*N了struct nope {    double x, y;}A[N];struct Dancing_Links_X {    void init() {        head = 0;        for (int i = 0; i <= n; i++) {            S[i] = 0;            mapp[i].up = mapp[i].down = i;            mapp[i].left = (i == 0 ? n : i - 1);            mapp[i].right = (i == n ? 0 : i + 1);        }        cnt = n;        memset(H, -1, sizeof(H));    }    void link(int x, int y) {        cnt++;        mapp[cnt].row = x;        mapp[cnt].col = y;        S[y]++;        mapp[cnt].up = mapp[y].up;        mapp[cnt].down = y;        mapp[mapp[y].up].down = cnt;        mapp[y].up = cnt;        if (H[x] == -1) H[x] = mapp[cnt].left = mapp[cnt].right = cnt;        else {            mapp[cnt].left = mapp[H[x]].left;            mapp[cnt].right = H[x];            mapp[mapp[H[x]].left].right = cnt;            mapp[H[x]].left = cnt;        }    }    void remove(int c) {//删去c这个点,以及关联的一列        for (int i = mapp[c].down; i != c; i = mapp[i].down) {            mapp[mapp[i].left].right = mapp[i].right;            mapp[mapp[i].right].left = mapp[i].left;        }    }    void resume(int c) {//恢复c这个点，以及关联的一列        for (int i = mapp[c].up; i != c; i = mapp[i].up) {            mapp[mapp[i].left].right = i;            mapp[mapp[i].right].left = i;        }    }    bool used[N];    int h() {        int sum = 0;        for (int i = mapp[head].right; i != head; i = mapp[i].right) used[i] = false;        for (int i = mapp[head].right; i != head; i = mapp[i].right) {            if (!used[i]) {                sum++;                used[i] = true;                for (int j = mapp[i].down; j != i; j = mapp[j].down)                    for (int k = mapp[j].right; k != j; k = mapp[k].right)                         used[mapp[k].col] = true;            }        }        return sum;    }    void dance(int nums) {        if (nums + h() >= len) return;        if (mapp[head].right == head) {            len = min(len, nums);            return;        }        int s = inf, c;        for (int t = mapp[head].right; t != head; t = mapp[t].right) {            if (S[t] < s) s = S[t], c = t;        }        for (int i = mapp[c].down; i != c; i = mapp[i].down) {            remove(i);            for (int j = mapp[i].right; j != i; j = mapp[j].right) {                remove(j);            }            dance(nums + 1);            for (int j = mapp[i].left; j != i; j = mapp[j].left) {                resume(j);            }            resume(i);        }        return;    }}DLX;double dis(int a, int b) {    return sqrt((A[a].x - A[b].x)*(A[a].x - A[b].x) + (A[a].y - A[b].y)*(A[a].y - A[b].y));}bool solve(int mid) {    DLX.init();    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= n; j++) {            if (lens[i][j]<=diss[mid]) {                DLX.link(j, i);            }        }    }    len = m + 1;    DLX.dance(0);    if (len > m) {        return false;    }    return true;}int main() {    int T;    int size;    scanf("%d", &T);    while (T--) {        size = 0;        scanf("%d%d", &n, &m);        for (int i = 1; i <= n; i++) scanf("%lf%lf", &A[i].x, &A[i].y);        for (int i = 1; i <= n; i++) {            for (int j = 1; j <= n; j++) {                lens[i][j] = dis(i, j);                diss[size++] = lens[i][j];            }        }        int l, r, mid, ans;        sort(diss, diss + size);        l = 0; r = unique(diss, diss + size) - diss - 1;        while (r >= l) {            mid = (l + r) / 2;            if (solve(mid)) r = mid - 1, ans = mid;            else l = mid + 1;        }        printf("%.6lf\n", diss[ans]);    }    return 0;}