hdu 2295 &&hdu 5046

来源：互联网发布：台式电脑连接不上网络编辑：程序博客网时间：2024/06/04 01:03

第一个题目的意思是有n个城市和m个雷达。你最多可以用k个雷达，问使用最小多少的半径可以使k个雷达覆盖n个城市。

第二个是九野要从n个城市选择k个城市建造机场，问最小的最大城市距离是多少

都是舞蹈链+剪枝+二分计算路径

贴第二题代码

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int M = 110;const int N = 30000;const int inf = 0x3f3f3f3f;const double ep = 1e-8;int k;struct DLX {    int n , m,size;    int U[N],D[N],R[N],L[N],row[N],col[N];    int ans[M],S[M];    int ansd;    void init(int a,int b) {        n = a;        m = b;        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++) ans[i] = -1;    }    void addRow(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(ans[r]<0)            ans[r] =  L[size] = R[size] = size;        else {            R[size] = R[ans[r]];            L[R[ans[r]]] = size;            L[size] = ans[r];            R[ans[r]] = size;        }    }    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]]=R[L[i]] = i;        }    }    bool v[N];    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) {        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]) {            remove(i);            for(int j=R[i]; j!=i; j=R[j])            remove(j);            if(dance(d+1)) return true;            for(int j=L[i]; j!=i; j=L[j])            resume(j);            resume(i);        }        return false;    }} g;struct node{   int x;   int y;}point[N],radar[N]; long long dis(node a,node b){     return (long long)abs(a.x-b.x)+(long long)abs(a.y-b.y);}int T;int n,m,p;int main(){    int cas = 0;scanf("%d",&T);while(T--){scanf("%d%d",&n,&k);for(int i=1;i<=n;i++){scanf("%d%d",&point[i].x,&point[i].y);}       /*        for(int i=1;i<=n;i++){            for(int j=1;j<=m;j++){                printf("%f ",dis(point[i],radar[j]));            }            puts("");        }        */        long long  ans = 0;long long  l= 0;long long  r = 100000000000LL;while(r >= l ){            //puts("OK");            g.init(n,n);            long long mid = (l+r)/2;            for(int i=1;i<=n;i++){                    for(int j=1;j<=n;j++){                    if(dis(point[i],point[j]) <= mid){                        g.addRow(j,i);                    }                    }            }            if(g.dance(0)) {r = mid - 1 ; ans = mid; }            else l = mid + 1 ;}printf("Case #%d: %I64d\n",++cas,ans);}}

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