HDU 2295 &&hdu 5046

The first topic means there are N cities and M radar. You can use a maximum of k radar and ask how much radius you can use to make the K radar cover n cities.

The second is the nine-wild to choose K cities from N cities to build an airport, ask the smallest maximum city distance is how much

It's all dance chains + pruning + Two-point computing path

Paste the second question code

#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,si    Ze    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", &AMP;POINT[I].X,&AMP;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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HDU 2295 &&hdu 5046