Moscow Subregional 2013.
比赛连接
http://opentrains.snarknews.info/~ejudge/team.cgi?contest_id=006570
总叙
一个铜牌队的题解
剩下的题，大概很久以后才会补了，太累了。。。。
A
队友做的，我吃瓜群众不知道
#include <bits/stdc++.h>
#define rep(a,b,c) for(int (a)=(b);(a)<=(c);++(a))
#define drep(a,b,c) for(int (a)=(b);(a)>=(c);--(a))
#define pb push_back
#define mp make_pair
#define sf scanf
#define pf printf
#define two(x) (1<<(x))
#define clr(x,y) memset((x),(y),sizeof((x)))
#define dbg(x) cout << #x << "=" << x << endl;
const int mod = 1e9 + 7;
int mul(int x,int y){return 1LL*x*y%mod;}
int qpow(int x , int y){int res=1;while(y){if(y&1) res=mul(res,x) ; y>>=1 ; x=mul(x,x);} return res;}
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}
using namespace std;
const int maxn = 1e4 + 15;
int K[maxn];
int main(int argc,char *argv[]){
int N=read();
rep(i,1,N) K[i]=read();
if( N < 3 ) pf("0\n");
else{
sort( K + 1 , K + N + 1 );
int mx = K[N] ;
long long sum = 0;
for(int i = 1 ; i < N ; ++ i) sum += K[i];
if( mx > 2LL * sum ) cout << sum << endl;
else cout << (1LL*mx + sum) / 3LL << endl;
}
return 0;
}
B
队友做的，吃瓜群众还是不知道
#include <bits/stdc++.h>
#define rep(a,b,c) for(int (a)=(b);(a)<=(c);++(a))
#define drep(a,b,c) for(int (a)=(b);(a)>=(c);--(a))
#define pb push_back
#define mp make_pair
#define sf scanf
#define pf printf
#define two(x) (1<<(x))
#define clr(x,y) memset((x),(y),sizeof((x)))
#define dbg(x) cout << #x << "=" << x << endl;
const int mod = 1e9 + 7;
int mul(int x,int y){return 1LL*x*y%mod;}
int qpow(int x , int y){int res=1;while(y){if(y&1) res=mul(res,x) ; y>>=1 ; x=mul(x,x);} return res;}
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}
using namespace std;
const int maxn = 20 + 15;
char str[maxn];
int len,vis[maxn][1<<16][2];
string dp[maxn][1<<16][2];
char number[16] = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
int use[20];
inline void up(string & x , string v){
if( x == "" ) x = v;
else x = min( x , v );
}
int main(int argc,char *argv[]){
//freopen("in.txt","r",stdin);
sf("%s",str+1);
len = strlen( str + 1 );
vis[0][0][0] = 1;
rep(i,0,len-1) rep(mask,0,two(16)-1) rep(f,0,1) if(vis[i][mask][f]){
int st = str[i + 1] <= '9' ? str[i + 1]-'0' : str[i + 1] - 'A' + 10;
if( f ) st = 0;
rep(add , st , 15 )
if( ( mask >> add & 1 ) == 0 ){
up( dp[i + 1][ mask | two( add ) ][ f | ( add > st ) ] , dp[i][mask][f] + number[add] );
vis[i + 1][mask | two( add )][ f | (add > st) ] = 1;
}
}
string ans = "GGGGGGGGGGGGGGGGGGGGGGGGGGG";
rep(mask,0,two(16)-1) if(vis[len][mask][1]) ans=min(ans,dp[len][mask][1]);
if( ans == "GGGGGGGGGGGGGGGGGGGGGGGGGGG" ){
rep(i,1,len+1){
int st = i == 1 ? 1 : 0;
rep(j,st,15){
if(!use[j]){
if( j <= 9 ) cout << j;
else cout << (j - 10) + 'A';
use[j] = 1;
break;
}
}
}
cout << endl;
}else cout << ans << endl;
return 0;
}
F
题意：有一个地方有两种交通工具，地下的需要28元，地上的需要26元，他可以花44元使得这90分钟随便坐车。
让你输出这个人如果贪心的去坐车，和最优策略去坐车的花费是多少。
题解：数据范围1500，所以DP就好了，这道题读题比做题难。。。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1500;
int n;
int t[maxn],op[maxn];
int s[3];
long long dp[maxn];
int main(){
s[1]=28;
s[2]=26;
scanf("%d",&n);
for(int i=1;i<=n;i++){
string s1,s2;
cin>>s1>>s2;
int now = 0;
for(int j=0;j<2;j++)
now = now*10 + s1[j]-'0';
int next = 0;
for(int j=3;j<s1.size();j++)
next = next*10 + s1[j]-'0';
t[i]=now*60+next;
if(s2[0]=='U')op[i]=1;
else op[i]=2;
}
t[n+1]=100000000;
long long ans1 = 0;
for(int i=1;i<=n;i++){
long long tmp = 0;
int flag = 0;
int j=i;
for(;j<=n;j++){
tmp+=s[op[j]];
if(op[j]==1)flag=1;
if(t[j+1]-t[i]>90)break;
if(op[j+1]==1&&flag)break;
}
if(tmp<44)ans1+=tmp;
else ans1+=44,i=j;
}
for(int i=1;i<=n;i++){
dp[i]=dp[i-1]+s[op[i]];
int flag = 0;
if(op[i]==1)flag=1;
for(int j=i-1;j>=1;j--){
if(flag&&op[j]==1)break;
if(op[j]==1)flag=1;
if(t[i]-t[j]>90)break;
dp[i]=min(dp[i],dp[j-1]+44);
}
//cout<<dp[i]<<endl;
}
cout<<ans1<<" "<<dp[n]<<endl;
}
H
这场比赛的签到水题
#include<bits/stdc++.h>
using namespace std;
int main(){
int x1,y1,x2,y2,x3,y3;
cin>>x1>>y1>>x2>>y2>>x3>>y3;
if(x1>x2)swap(x1,x2);
if(y1>y2)swap(y1,y2);
if(x1<=x3&&x3<=x2&&y1<=y3&&y3<=y2)puts("Yes");
else puts("No");
}
I
题意：有n个集合，然后给你两两集合共有的元素，问你能不能构造出来这些集合，数据范围200
题解：直观的想法就是共有的就直接插进去，然后最后check一下就好了。我感觉直接暴力是n^4的，所以我bitset了一下。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 205;
int v1[maxn*maxn],v2[maxn*maxn];
vector<int> v3[maxn*maxn];
bitset<10005>S[205];
bitset<10005>tmp;
int main(){
int n;
scanf("%d",&n);
for(int i=1;i<=n*(n-1)/2;i++){
scanf("%d%d",&v1[i],&v2[i]);
int num;scanf("%d",&num);
for(int j=1;j<=num;j++){
int x;scanf("%d",&x);
S[v1[i]][x]=1;
S[v2[i]][x]=1;
v3[i].push_back(x);
}
sort(v3[i].begin(),v3[i].end());
}
int flag = 0;
for(int i=1;i<=n*(n-1)/2;i++){
tmp = S[v1[i]]&S[v2[i]];
if(tmp.count()>v3[i].size()){
printf("No\n");
return 0;
}
}
printf("Yes\n");
for(int i=1;i<=n;i++){
printf("%d",S[i].count());
for(int j=0;j<10005;j++){
if(S[i][j])printf(" %d",j);
}
printf("\n");
}
}
K
题意：给你x[-1],x[0],A,B,C,然后X[i] = (AX[i-2] + BX[i] + C )%2^32
然后让你算出[1,n]的xi值之后，输出前k大的，n<=1e8,k<=2e5
题解：这种题一看就是瞎JB莽，我们队分情况讨论了很多种……
但是似乎直接前面暴力2e5个数，后面暴力2e5个，再for一遍就好了。。。
尴尬。
#include<bits/stdc++.h>
using namespace std;
const int mod = 0x7fffffff;
vector<int> pppp;
int x0,x1,A,B,C;
int get(){
int a = (A*x0+B*x1+C)&(mod);
return a;
}
int n,k;
struct Heap
{
const static int HeapSize = 2e5 + 500;
int val[HeapSize] , sz;
//向上维护
void maintain_up(int pos){
int cur = pos , pre = cur >> 1;
while(pre){
if(val[cur] < val[pre]) swap(val[cur] , val[pre]) , cur = pre;
else break;
pre = cur >> 1;
}
}
//向下维护
void maintain_down(int pos){
int cur = pos;
while(cur * 2 <= sz){
int lson = cur << 1;
int rson = cur << 1 | 1;
if(rson > sz) rson ^= 1;
int nxt = lson;
if(val[rson] < val[lson]) nxt = rson;
if(val[cur] > val[nxt]) swap(val[cur],val[nxt]) , cur = nxt;
else break;
}
}
void insert(int x){
val[++sz] = x;
maintain_up(sz);
}
int top(){
return val[1];
}
void pop(){
if(sz==0) return;
swap(val[1] , val[sz--]);
maintain_down( 1 );
}
void init(){ sz = 0 ;}
}heap;
void solve1(){
heap.init();
int tmp = get();
x0=tmp,swap(x0,x1);
heap.insert(tmp);
int now = heap.top();
for(int i=2;i<=k;i++){
int tmp=get();
heap.insert(tmp);
now=heap.top();
x0=tmp,swap(x0,x1);
}
for(int i=k+1;i<=n;i++){
int tmp = get();
if(now<tmp){
heap.pop();
heap.insert(tmp);
now=heap.top();
}
x0=tmp,swap(x0,x1);
}
vector<int> ans;
for(int i=1;i<=k;i++){
ans.push_back(heap.top());
heap.pop();
}
reverse(ans.begin(),ans.end());
for(int i=0;i<ans.size();i++)
printf("%d ",ans[i]);
printf("\n");
}
void solve2(){
for(int i=1;i<=k;i++)
printf("%d ",C);
printf("\n");
}
int get1(int x){
return (x1+C*x)&(mod);
}
void solve3(){
heap.init();
int num = min(200000,n);
for(int i=1;i<=min(200000,n);i++)
pppp.push_back(i);
for(int i=n;i>=max(1,n-200000);i--)
pppp.push_back(i);
sort(pppp.begin(),pppp.end());
pppp.erase(unique(pppp.begin(),pppp.end()),pppp.end());
int tmp = get1(pppp[0]);
heap.insert(tmp);
int now = heap.top();
for(int i=1;i<pppp.size();i++){
int tmp=get1(pppp[i]);
if( heap.sz < k ){
heap.insert( tmp );
now = heap.top();
}else if(now<tmp){
heap.pop();
heap.insert(tmp);
now=heap.top();
}
}
int j = 0;
reverse(pppp.begin(),pppp.end());
for(int i=n;i>=1;--i){
while(j<pppp.size()&&pppp[j]>i) ++ j;
if( j < pppp.size() && pppp[j] == i) continue;
int tmp = get1(i);
if( heap.sz < k ){
heap.insert( tmp );
now = heap.top();
}
else if(now<tmp){
heap.pop();
heap.insert(tmp);
now=heap.top();
}
}
vector<int> ans;
for(int i=1;i<=k;i++){
ans.push_back(heap.top());
heap.pop();
}
reverse(ans.begin(),ans.end());
for(int i=0;i<ans.size();i++)
printf("%d ",ans[i]);
printf("\n");
}
int get2(int x){
if(x&1)return (x0+(x+1)/2*C)&(mod);
else return (x1+x/2*C)&(mod);
//return (x0+C*x)&(mod);
}
void solve4(){
heap.init();
int num = min(200000,n);
for(int i=1;i<=min(200000,n);i++)
pppp.push_back(i);
for(int i=n;i>=max(1,n-200000);i--)
pppp.push_back(i);
sort(pppp.begin(),pppp.end());
pppp.erase(unique(pppp.begin(),pppp.end()),pppp.end());
int tmp = get2(pppp[0]);
heap.insert(tmp);
int now = heap.top();
for(int i=1;i<pppp.size();i++){
int tmp=get2(pppp[i]);
if( heap.sz < k ){
heap.insert( tmp );
now = heap.top();
}else if(now<tmp){
heap.pop();
heap.insert(tmp);
now=heap.top();
}
}
int j = 0;
reverse(pppp.begin(),pppp.end());
for(int i=n;i>=1;--i){
while(j<pppp.size()&&pppp[j]>i) ++ j;
if( j < pppp.size() && pppp[j] == i) continue;
int tmp = get2(i);
if( heap.sz < k ){
heap.insert( tmp );
now = heap.top();
}
else if(now<tmp){
heap.pop();
heap.insert(tmp);
now=heap.top();
}
}
vector<int> ans;
for(int i=1;i<=k;i++){
ans.push_back(heap.top());
heap.pop();
}
reverse(ans.begin(),ans.end());
for(int i=0;i<ans.size();i++)
printf("%d ",ans[i]);
printf("\n");
}
int main(){
//freopen("a+b.in","r",stdin);
// freopen("out.txt","w",stdout);
srand(time(NULL));
scanf("%d%d",&n,&k);
cin>>x0>>x1>>A>>B>>C;
if(A==0&&B==0)solve2();
else if(A==0&&B==1)solve3();
else if(A==1&&B==0)solve4();
else if(A>=1||B>=1)solve1();
return 0;
}