HDU 2295 Radar (DLX求重复覆盖, A*搜索)

题目大意：

就是在M个站中选择至多K个使得N个点被覆盖, 为需要的最小半径

大致思路：

二分答案R, 然后建立N*M的01矩阵判断重复覆盖是否可行

重复覆盖和精确覆盖不同, 每次只会删掉每一列所有相关的1, 而不会将有相关1的行删去, 所以矩阵在减少的速度上没有精确覆盖快, 需要进行剪枝

这里使用一个估价函数f(), 表示单签状况下最好情况需要多少步才能走完, 进行剪枝

代码如下：

Result  :  Accepted     Memory  :  1656 KB     Time  :  62 ms

/* * Author: Gatevin * Created Time:  2015/10/4 18:10:46 * File Name: Sakura_Chiyo.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define maxnode 4000#define maxn 60#define maxm 60pair<int, int> city[60];pair<int, int> sta[60];struct DLX{    int n, m, size;    int U[maxnode], D[maxnode], R[maxnode], L[maxnode], Row[maxnode], Col[maxnode];    int H[maxn], S[maxm];    int ansd, ans[maxn];    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;            L[i] = i - 1;            R[i] = i + 1;        }        R[m] = 0; L[0] = m;        size = m;        for(int i = 1; i <= n; i++) H[i] = -1;    }    void Link(int r, int c)    {        ++S[Col[++size] = c];        Row[size] = r;        D[size] = D[c];        U[D[c]] = size;        U[size] = c;        D[c] = size;        if(H[r] < 0) H[r] = L[size] = R[size] = size;        else        {            R[size] = R[H[r]];            L[R[H[r]]] = size;            L[size] = H[r];            R[H[r]] = size;        }    }    void remove(int c)//因为是重复覆盖每次讲某一列的所有1纳入不考虑范围    {        //L[R[c]] = L[c]; R[L[c]] = R[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;        //L[R[c]] = R[L[c]] = c;    }    bool v[60];    int f()//A*搜索估价函数    {        int ret = 0;        for(int i = R[0]; i != 0; i = R[i]) v[i] = 1;        for(int i = R[0]; i != 0; i = R[i])            if(v[i])            {                ret++;                v[i] = 0;                for(int j = D[i]; j != i; j = D[j])                    for(int k = R[j]; k != j; k = R[k])                        v[Col[k]] = 0;            }        return ret;    }        int Dance(int dep, int K)    {        if(dep + f() > K) return false;        if(R[0] == 0)        {            return true;        }        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])        {            for(int j = R[i]; j != i; j = R[j]) remove(j);            remove(i);            if(Dance(dep + 1, K)) return true;            resume(i);            for(int j = L[i]; j != i; j = L[j]) resume(j);        }        return false;    }    void solve()    {        int N, M, K;        scanf("%d %d %d", &N, &M, &K);        for(int i = 1; i <= N; i++)            scanf("%d %d", &city[i].first, &city[i].second);        for(int j = 1; j <= M; j++)            scanf("%d %d", &sta[j].first, &sta[j].second);                double L = 0, R = 1500, mid = -1, ans = -1;        int cnt = 0;        int times = 40;        while(L + eps < R && cnt <= times)        {            cnt++;            mid = (L + R) / 2.;            init(M, N);            for(int i = 1; i <= M; i++)                for(int j = 1; j <= N; j++)                    if((sta[i].first - city[j].first)*(sta[i].first - city[j].first)*1.                            + (sta[i].second - city[j].second)*(sta[i].second - city[j].second)*1. < mid*mid)                        Link(i, j);            if(Dance(0, K))            {                ans = mid;                R = mid;            }            else L = mid;        }        printf("%.6f\n", ans);        return;    }};DLX dlx;int main(){    int T;    scanf("%d", &T);    while(T--)        dlx.solve();    return 0;}