P4799 [CEOI2015 Day2]世界冰球锦标赛
考察:双向dfs
思路:
爆搜的时间复杂度是O(240).明显TLE.背包的时间复杂度是O(m*n) 也会TLE.所以需要优化.
这里的优化考虑折半搜索.也就只搜索一半.在搜索另一半的时候进行答案累加.这里先预处理在0~n/2的比赛里选的方案.然后在搜另一半的时候二分找到最大的与当前花费和<=m的左半边下标.这样就优化查找.具体还是看代码吧...
本蒟蒻自己写的代码T了...O2能过,但还是参考题解不开O2.题解是预处理了两边的方案再累加.
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <vector> 5 using namespace std; 6 typedef long long LL; 7 typedef pair<LL,int> PIL; 8 const int N = 45,M = 1<<23; 9 LL c[N],m,ans,L[M],R[M]; 10 int n,sz; 11 void dfs(int st,LL cost,LL a[],int ed,int& cnt) 12 { 13 if(st>ed) 14 { 15 a[cnt++] = cost; 16 return; 17 } 18 if(c[st]<=m-cost) dfs(st+1,cost+c[st],a,ed,cnt); 19 dfs(st+1,cost,a,ed,cnt); 20 } 21 int main() 22 { 23 scanf("%d%lld",&n,&m); 24 for(int i=0;i<n;i++) scanf("%lld",&c[i]); 25 sort(c,c+n); 26 reverse(c,c+n); 27 int cnt = 0; 28 dfs(0,0,L,n/2,sz); 29 dfs(n/2+1,0,R,n-1,cnt); 30 sort(L,L+sz); 31 for(int i=0;i<cnt;i++) 32 { 33 LL idx = upper_bound(L,L+sz,m-R[i])-L; 34 ans+=idx; 35 } 36 printf("%lld\n",ans); 37 return 0; 38 }
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 typedef long long LL; 6 typedef pair<LL,int> PIL; 7 const int N = 45,M = 1<<22; 8 LL c[N],m,ans; 9 int n,sz; 10 PIL l[M]; 11 void dfs_1(int st,int now,LL cost) 12 { 13 l[sz++] = {cost,now}; 14 if(cost+c[n/2]>m) return; 15 if(st>n/2) return; 16 if(!(now>>st&1)&&c[st]<=m-cost) dfs_1(st+1,now|(1<<st),cost+c[st]); 17 dfs_1(st+1,now,cost); 18 } 19 void dfs_2(int st,int now,LL cost) 20 { 21 int low = 0,high = sz-1; 22 while(low<high) 23 { 24 int mid = low+high+1>>1; 25 if(l[mid].first<=m-cost) low = mid; 26 else high = mid-1; 27 } 28 if(high>=0) ans+=high+1; 29 for(int i=st;i<n;i++) 30 if(!(now>>st&1)&&c[i]<=m-cost) dfs_2(i+1,now|(1<<i),cost+c[i]); 31 } 32 int main() 33 { 34 scanf("%d%lld",&n,&m); 35 for(int i=0;i<n;i++) scanf("%lld",&c[i]); 36 sort(c,c+n); 37 reverse(c,c+n); 38 dfs_1(0,0,0); 39 sort(l,l+sz); 40 sz = unique(l,l+sz)-l; 41 dfs_2(n/2+1,0,0); 42 printf("%lld\n",ans); 43 return 0; 44 }
