Codeforces Round #541 (Div. 2)
http://codeforces.com/contest/1131
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 1000005 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 14 int main(){ 15 #ifndef ONLINE_JUDGE 16 // freopen("input.txt","r",stdin); 17 #endif 18 std::ios::sync_with_stdio(false); 19 ll w1,h1,w2,h2; 20 cin>>w1>>h1>>w2>>h2; 21 ll tmp1=(w2+2)*(h2+2);//上面的 22 ll tmp2=(w1+2)*(h1+2); 23 ll tmp3=w1*h1+w2*h2; 24 ll ans=tmp1+tmp2-tmp3-2*(w2+2); 25 cout<<ans<<endl; 26 }
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 1000005 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 int main(){ 14 #ifndef ONLINE_JUDGE 15 // freopen("input.txt","r",stdin); 16 #endif 17 std::ios::sync_with_stdio(false); 18 int n; 19 cin>>n; 20 ll a,b,x,y; 21 x=0,y=0; 22 ll ans=1; 23 for(int i=1;i<=n;i++){ 24 cin>>a>>b; 25 ans+=max(0LL,min(a,b)-max(x,y)+(x!=y)); 26 x=a,y=b; 27 } 28 cout<<ans<<endl; 29 }
C
sort排序，然后从小到大依次左右各添一个向中间添加。。。在比赛的时候第一感觉这就是正解，但是不会证明为什么。。。
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 1000005 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 int a[105]; 14 int ans[105]; 15 16 int main(){ 17 #ifndef ONLINE_JUDGE 18 // freopen("input.txt","r",stdin); 19 #endif 20 std::ios::sync_with_stdio(false); 21 int n; 22 cin>>n; 23 for(int i=1;i<=n;i++) cin>>a[i]; 24 sort(a+1,a+n+1); 25 int L=1,R=n; 26 int i=1; 27 while(L<=R){ 28 if(i%2) 29 ans[L++]=a[i]; 30 else 31 ans[R--]=a[i]; 32 i++; 33 } 34 for(int i=1;i<=n;i++) cout<<ans[i]<<" "; 35 }
D
用并查集把相同的点连在一起，然后剩下的点用拓扑排序即可
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 2005 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 int n,m; 14 char str[1005][1005]; 15 16 int fa[maxn]; 17 int d[maxn]; 18 int vis[maxn]; 19 int ans[maxn]; 20 vector<int>ve[maxn]; 21 22 int Find(int x){ 23 int r=x,y; 24 while(x!=fa[x]){ 25 x=fa[x]; 26 } 27 while(r!=x){ 28 y=fa[r]; 29 fa[r]=x; 30 r=y; 31 } 32 return x; 33 } 34 35 void join(int x,int y){ 36 int xx=Find(x); 37 int yy=Find(y); 38 fa[xx]=yy; 39 } 40 41 void add(int x,int y){ 42 int xx=Find(x); 43 int yy=Find(y); 44 if(xx!=yy){ 45 d[xx]++; 46 ve[yy].pb(xx); 47 } 48 else{ 49 cout<<"NO"<<endl; 50 exit(0); 51 } 52 } 53 54 void topsort(){ 55 queue<pair<int,int> >Q; 56 rep(i,1,n+m+1){ 57 int xx=Find(i); 58 if(!d[xx]&&!vis[xx]){ 59 vis[xx]=1; 60 Q.push(make_pair(xx,1)); 61 } 62 } 63 pair<int,int>pp; 64 while(!Q.empty()){ 65 pp=Q.front(); 66 Q.pop(); 67 int pos=pp.first; 68 int v=pp.second; 69 ans[pos]=v; 70 rep(i,0,ve[pos].size()){ 71 int w=ve[pos][i]; 72 d[w]--; 73 if(!d[w]){ 74 Q.push(make_pair(w,v+1)); 75 } 76 } 77 } 78 rep(i,1,n+m+1){ 79 if(d[i]){ 80 cout<<"NO"<<endl; 81 exit(0); 82 } 83 } 84 } 85 86 int main(){ 87 #ifndef ONLINE_JUDGE 88 freopen("input.txt","r",stdin); 89 #endif 90 std::ios::sync_with_stdio(false); 91 cin>>n>>m; 92 rep(i,1,n+m+1) fa[i]=i; 93 rep(i,1,n+1) cin>>(str[i]+1); 94 rep(i,1,n+1){ 95 rep(j,1,m+1){ 96 if(str[i][j]=='='){ 97 join(Find(i),Find(j+n)); 98 } 99 } 100 } 101 rep(i,1,n+1){ 102 rep(j,1,m+1){ 103 if(str[i][j]=='>'){ 104 add(i,n+j); 105 } 106 else if(str[i][j]=='<'){ 107 add(n+j,i); 108 } 109 } 110 } 111 topsort(); 112 cout<<"YES"<<endl; 113 rep(i,1,n+1) cout<<ans[Find(i)]<<" "; 114 cout<<endl; 115 rep(i,1,m+1) cout<<ans[Find(i+n)]<<" "; 116 }
E
类似模拟的方法
找出字符串的连续前缀，连续后缀和中间最长连续子串
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 1000005 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 int ans=0; 14 string s[100005]; 15 16 int main(){ 17 #ifndef ONLINE_JUDGE 18 // freopen("input.txt","r",stdin); 19 #endif 20 std::ios::sync_with_stdio(false); 21 int n; 22 cin>>n; 23 for(int i=0;i<n;i++){ 24 cin>>s[i]; 25 } 26 for(int c=0;c<26;c++){ 27 int l=0,r=0,all=1; 28 for(int i=n-1;i>=0;i--){ 29 int cnt=0,pf=0,sf=0,tt=0; 30 while(pf<s[i].length()&&s[i][pf]=='a'+c) ++pf; 31 while(sf<s[i].length()&&s[i][s[i].length()-1-sf]=='a'+c) ++sf; 32 for(int j=0;j<s[i].length();j++){ 33 if(s[i][j]=='a'+c){ 34 cnt=max(cnt,++tt); 35 } 36 else{ 37 tt=0; 38 } 39 } 40 if(all){ 41 ans=max(ans,(cnt+1)*(l+1)-1); 42 l=(pf+1)*(l+1)-1; 43 r=(sf+1)*(r+1)-1; 44 if(pf!=s[i].length()){ 45 all=0; 46 } 47 } 48 else{ 49 if(cnt){ 50 ans=max(ans,l+r+1); 51 } 52 } 53 } 54 } 55 cout<<ans<<endl; 56 }
F
比赛的时候犯傻了，一直在想建图，然后找度为1的点，然后深搜的错误性。。没想出来。后来才知道了错误性
正解应该是：并查集，合并就完事了
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 1000005 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 vector<int>ve[150005]; 14 15 int fa[150005]; 16 17 int Find(int x){ 18 int r=x,y; 19 while(x!=fa[x]){ 20 x=fa[x]; 21 } 22 while(r!=x){ 23 y=fa[r]; 24 fa[r]=x; 25 r=y; 26 } 27 return x; 28 } 29 30 31 void join(int x,int y){ 32 int xx=Find(x); 33 int yy=Find(y); 34 if(ve[xx].size()>ve[yy].size()){ 35 swap(xx,yy); 36 } 37 rep(i,0,ve[xx].size()){ 38 ve[yy].pb(ve[xx][i]); 39 fa[ve[xx][i]]=yy; 40 } 41 } 42 43 int main(){ 44 #ifndef ONLINE_JUDGE 45 freopen("input.txt","r",stdin); 46 #endif 47 std::ios::sync_with_stdio(false); 48 int n; 49 cin>>n; 50 rep(i,1,n+1) fa[i]=i,ve[i].pb(i); 51 int u,v; 52 rep(i,1,n){ 53 cin>>u>>v; 54 join(u,v); 55 } 56 int pos=Find(1); 57 rep(i,0,ve[pos].size()){ 58 cout<<ve[pos][i]<<" "; 59 } 60 61 }