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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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原文地址:http://blog.csdn.net/u013076044/article/details/46897675