Codeforces Beta Round #72 (Div. 2 Only)
http://codeforces.com/contest/84
A
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef unsigned long long ull; 14 int main(){ 15 #ifndef ONLINE_JUDGE 16 // freopen("input.txt","r",stdin); 17 #endif 18 std::ios::sync_with_stdio(false); 19 ll n; 20 cin>>n; 21 cout<<2*n-n/2<<endl; 22 }
B
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef unsigned long long ull; 14 15 int n; 16 ll a[100005]; 17 18 int main(){ 19 #ifndef ONLINE_JUDGE 20 // freopen("input.txt","r",stdin); 21 #endif 22 std::ios::sync_with_stdio(false); 23 cin>>n; 24 for(int i=1;i<=n;i++) cin>>a[i]; 25 ll ans=n; 26 ll co=1; 27 for(int i=1;i<n;i++){ 28 if(a[i]==a[i+1]){ 29 co++; 30 } 31 else{ 32 ans+=co*(co-1)/2; 33 co=1; 34 } 35 } 36 ans+=co*(co-1)/2; 37 cout<<ans<<endl; 38 }
C
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef unsigned long long ull; 14 15 const int N = 10010, D = 20000; 16 int n, m, x, y, c[N], r[N], s[N]; 17 vector<int> v[D * 2 + 10]; 18 19 int main(){ 20 #ifndef ONLINE_JUDGE 21 // freopen("input.txt","r",stdin); 22 #endif 23 std::ios::sync_with_stdio(false); 24 cin>>n; 25 int ans=0; 26 for (int i = 1; i <= n; i++){ 27 cin>>c[i]>>r[i]; 28 c[i] += D; s[i] = -1; 29 for (int j = c[i] - r[i]; j <= c[i] + r[i]; j++) v[j].push_back(i); 30 } 31 cin>>m; 32 for (int i = 1; i <= m; i++){ 33 cin>>x>>y; 34 x += D; 35 for (auto j : v[x]) if (s[j] == -1) 36 if (y * y + (x - c[j]) * (x - c[j]) <= r[j] * r[j]){ 37 s[j] = i; 38 ans++; 39 } 40 } 41 cout<<ans<<endl; 42 for (int i = 1; i <= n; i++) cout<<s[i]<<" "; 43 }
D
二分+模拟
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef unsigned long long ull; 14 15 ll a[100005]; 16 ll n,k; 17 int book[100005]; 18 vector<int>ans; 19 20 bool erfen(ll mid){ 21 ll sum=0; 22 for(int i=1;i<=n;i++){ 23 sum+=min(a[i],mid); 24 } 25 if(sum>=k) return true; 26 return false; 27 } 28 29 int main(){ 30 #ifndef ONLINE_JUDGE 31 // freopen("input.txt","r",stdin); 32 #endif 33 std::ios::sync_with_stdio(false); 34 cin>>n>>k; 35 ll sum=0; 36 ll Max=0; 37 ll L,R,mid; 38 for(int i=1;i<=n;i++){ 39 cin>>a[i]; 40 sum+=a[i]; 41 Max=max(Max,a[i]); 42 } 43 if(sum<k) cout<<-1<<endl; 44 else if(sum==k){} 45 else{ 46 L=0,R=Max,mid; 47 while(L<=R){ 48 mid=L+R>>1; 49 if(erfen(mid)){ 50 R=mid-1; 51 } 52 else{ 53 L=mid+1; 54 } 55 } 56 ll kk=R; 57 sum=0; 58 for(int i=1;i<=n;i++){ 59 sum+=min(a[i],kk); 60 a[i]-=min(a[i],kk); 61 } 62 int flag=1; 63 for(int i=1;i<=n;i++){ 64 if(sum>k&&a[i]>0){ 65 book[i]=1; 66 ans.pb(i); 67 } 68 else if(a[i]>0){ 69 sum++; 70 a[i]--; 71 if(sum>k&&flag&&a[i]>=0){ 72 flag=0; 73 ans.pb(i); 74 book[i]=1; 75 } 76 } 77 78 } 79 for(int i=1;i<=n;i++){ 80 if(!book[i]&&a[i]>0){ 81 book[i]=1; 82 ans.pb(i); 83 } 84 } 85 for(int i=0;i<ans.size();i++){ 86 cout<<ans[i]<<" "; 87 } 88 } 89 90 }
E
__builtin_popcount() 函数，计算一个数转二进制之后有几个1
一道很好的最短路的题目。
题意：求S到T的最短路径，其中，路径上不同的字母不能超过K个，且答案的字典序要最小，不存在输出-1
思路：因为K小于等于4,所以可以用状压来存放不同字母的个数。在优先队列中存放距离，路径的字符串，K和该点的下表，跑一次最短路即可。
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef pair<int,int> pii; 14 typedef pair<pair<int,string>,pii> ppp; 15 typedef unsigned long long ull; 16 17 int n,m,k; 18 int S,T; 19 string s[55]; 20 set<pii>se; 21 22 int dist(int a,int b){ 23 return abs(a/m-b/m)+abs(a%m-b%m); 24 } 25 26 int dir[4][2]={0,1,1,0,0,-1,-1,0}; 27 28 void Dijstra(){ 29 priority_queue<ppp,vector<ppp>,greater<ppp>>Q; 30 Q.push({{dist(S,T),""},{0,S}}); 31 ppp ss; 32 string str; 33 int book,pos; 34 int x,y,xx,yy; 35 while(!Q.empty()){ 36 ss=Q.top(); 37 Q.pop(); 38 str=ss.first.second; 39 book=ss.second.first; 40 pos=ss.second.second; 41 if(!se.count({book,pos})){ 42 se.insert({book,pos}); 43 x=pos/m,y=pos%m; 44 for(int i=0;i<4;i++){ 45 xx=x+dir[i][0]; 46 yy=y+dir[i][1]; 47 if(xx>=0&&xx<n&&yy>=00&&yy<m){ 48 if(s[xx][yy]=='T'){ 49 cout<<str<<endl; 50 exit(0); 51 } 52 if(__builtin_popcount(book|(1<<(s[xx][yy]-'a')))<=k) 53 Q.push({{dist(xx*m+yy,T)+str.length()+1,str+s[xx][yy]},{book|(1<<(s[xx][yy]-'a')),xx*m+yy}}); 54 } 55 } 56 } 57 } 58 59 } 60 61 int main(){ 62 #ifndef ONLINE_JUDGE 63 // freopen("input.txt","r",stdin); 64 #endif 65 std::ios::sync_with_stdio(false); 66 cin>>n>>m>>k; 67 for(int i=0;i<n;i++){ 68 cin>>s[i]; 69 for(int j=0;j<m;j++){ 70 if(s[i][j]=='S'){ 71 S=i*m+j; 72 } 73 if(s[i][j]=='T'){ 74 T=i*m+j; 75 } 76 } 77 } 78 Dijstra(); 79 cout<<-1<<endl; 80 }