10月21日 訓練記錄 GYM 101908
題意:博弈,二維平面上n個點,每次可以下移,左移或對角線移動任意步,將任意點移到原點即勝利
思路:SG函式,初始有在對角線上的點則先手必勝。有三個禁區x軸,y軸和對角線,一旦移動到這些地方對手就贏了。所以相應得(1,2)和(2,1)就成為必敗點了,這兩個點無論如何移動都會進入到禁區。任意點最終都會移到(1,2)和(2,1),所以以(1,2)和(2,1)為起點並躲避禁區SG博弈就出結果了
程式碼:
#include<bits/stdc++.h> using namespace std; const int maxn = 110; int sg[maxn][maxn],vis[510]; int main() { sg[1][2] = 0; sg[2][1] = 0; for ( int i=1 ; i<=100 ; i++ ) for ( int j=1 ; j<=100 ; j++ ) { if ( i==j ) continue; if ( i==1&&j==2 ) continue; if ( i==2&&j==1 ) continue; for ( int k=0 ; k<510 ; k++ ) vis[k] = 0; for ( int k=1 ; k<i ; k++ ) { if ( k==j ) continue; vis[sg[k][j]] = 1; } for ( int k=1 ; k<j ; k++ ) { if ( i==k ) continue; vis[sg[i][k]] = 1; } for ( int k=1 ; k<i&&k<j ; k++ ) vis[sg[i-k][j-k]] = 1; for ( int k=0 ; k<510 ; k++ ) { if ( !vis[k] ) { sg[i][j] = k; break; } } } for ( int n ; scanf( "%d" , &n )==1 ; ) { int ans = 0; bool win = false; for( int i=1 ; i<=n ; i++ ) { int l,c; scanf( "%d%d" , &l , &c ); if ( l==c ) win = true; else ans = ans^sg[l][c]; } if ( ans!=0||win ) printf("Y\n"); else printf("N\n"); } return 0; }
題意:切割一個X*Y的矩形,保證割線滿足
1.兩條割線最多隻有一個交點
2.三條割線不會交於一點
3.兩條割線的交點不會交在邊界線上
4.任意割線不會割在角落上
最終矩形最終被分割成多少塊
思路:由題意不難推出此題求的就是割線交點數+割線數,在不考慮垂直線互相相交和平行線互相相交的情況可得答案為(V+1)*(H+1)。計算垂直線和平行線互相相交用樹狀陣列統計即可
程式碼:
#include<bits/stdc++.h> using namespace std; typedef long long LL; const int maxn = 100010; int a[maxn],bit[maxn],m; struct node { int x,y; friend bool operator<( const node&a , const node&b ) { return a.x>b.x; } }Data[maxn]; int lowbit( int x ) { return x&(-x); } void add( int x , int val ) { for ( ; x<=m ; x+=lowbit(x) ) bit[x] += val; } int sum( int x ) { int res = 0; for ( ; x>=1 ; x-=lowbit(x) ) res += bit[x]; return res; } LL slove( int n ) { for ( int i=1 ; i<=n ; i++ ) scanf( "%d%d" , &Data[i].x , &Data[i].y ),a[i] = Data[i].y; sort ( a+1 , a+n+1 ); m = unique( a+1 , a+n+1 )-a-1; sort ( Data+1 , Data+n+1 ); LL res = 0; for ( int i=1 ; i<=m ; i++ ) bit[i] = 0; for ( int i=1 ; i<=n ; i++ ) { int t = lower_bound( a+1 , a+m+1 , Data[i].y )-a; res += sum( t ); add( t , 1 ); } return res; } int main() { for( int X,Y ; scanf( "%d%d" , &X , &Y )==2 ; ) { int H,V; scanf( "%d%d" , &H , &V ); LL ans = 1LL*(H+1)*(V+1); ans += slove( H ); ans += slove( V ); printf( "%lld\n" , ans ); } return 0; }
題意:統計非1的個數
思路:水題
程式碼:
#include<bits/stdc++.h> using namespace std; int main() { for ( int n ; scanf( "%d" , &n )==1 ; ) { int ans = 0; for ( int i=1 ; i<=n ; i++ ) { int x; scanf( "%d" , &x ); if ( x!=1 ) ans++; } printf( "%d\n" , ans ); } return 0; }
題意:給兩字串S,T,求S裡有多少長度為LEN(T)的子串和沒有一個字元相同
思路:此題複雜度為LEN(T)*(LEN(S)-LEN(T)+1)且LEN(S)<=10000,最大複雜度為O(5000*5001)
程式碼:
#include<bits/stdc++.h>
using namespace std;
char s[10010],t[10010];
int main()
{
scanf( "%s" , s );
scanf( "%s" , t );
int ans = 0;
int lens = strlen(s);
int lent = strlen(t);
for ( int i=0 ; i<=lens-lent ; i++ )
{
bool ok = true;
for ( int j=0 ; j<lent ; j++ )
if ( s[i+j]==t[j] )
{
ok = false;
break;
}
if ( ok ) ans++;
}
printf( "%d\n" , ans );
return 0;
}
題意:有n個場地,每個場地上有一些表演,要求每個場地都至少看一次表演,求看錶演是可以獲得最大價值
思路:n<=10,表演總數<=1000,狀壓DP:i表示時間節點,j表示狀態(表示那些場地去沒去過)
程式碼:
#include<bits/stdc++.h>
using namespace std;
int Max( int a , int b ){ return a>b?a:b; }
int n,N,m;
struct node
{
int l,r,val,idx;
friend bool operator<( const node&a , const node&b )
{
return a.r<b.r;
}
}a[1010];
int b[2010],dp[2010][1050];
int main()
{
scanf( "%d" , &n );
m = 0;
N = 0;
for ( int i=0 ; i<n ; i++ )
{
int k;
scanf( "%d" , &k );
for ( int j=0 ; j<k ; j++ )
{
scanf ( "%d%d%d" , &a[N].l , &a[N].r , &a[N].val );
a[N].idx = i;
b[m] = a[N].l; m++;
b[m] = a[N].r; m++;
N++;
}
}
sort ( b , b+m );
m = unique( b , b+m )-b;
/*
printf( "%d\n" , m );
for ( int i=0 ; i<m ; i++ )
printf( "%d " , b[i] );
printf( "\n" );
*/
for ( int i=0 ; i<2010 ; i++ )
for ( int j=0 ; j<1050 ; j++ )
dp[i][j] = -1;
dp[0][0] = 0;
sort ( a , a+N );
for ( int i=0 ; i<N ; i++ )
{
a[i].l = lower_bound( b , b+m , a[i].l )-b+1;
a[i].r = lower_bound( b , b+m , a[i].r )-b+1;
//printf ( "%d %d %d\n" , a[i].l , a[i].r , a[i].val );
}
for ( int i=0,lr=0 ; i<N ; i++ )
{
int L=i,R=i;
while ( R+1<N&&a[R+1].r==a[L].r ) R++;
for ( int j=lr+1 ; j<=a[L].r ; j++ )
for ( int k=0 ; k<(1<<n) ; k++ )
dp[j][k] = dp[j-1][k];
//printf ( "%d %d\n" , L , R );
for ( int j=L ; j<=R ; j++ )
{
for( int k=0 ; k<(1<<n) ; k++ )
if ( dp[a[j].l][k]!=-1 ) dp[a[j].r][k|(1<<a[j].idx)] = Max( dp[a[j].r][k|(1<<a[j].idx)] , dp[a[j].l][k]+a[j].val );
}
lr = a[L].r;
i = R;
}
/*
for ( int i=0 ; i<=m ; i++ )
{
for ( int j=0 ; j<1<<n ; j++ )
printf ( "%d " , dp[i][j] );
printf ( "\n" );
}
*/
printf ( "%d\n" , dp[m][(1<<n)-1] );
return 0;
}
題意:煉油廠往加油站運送汽油,有很多道路可以選擇不同道路花費的時間不同,求為所有加油站加滿所需的最短時間
思路:二分時間,壓時間建圖,最大流看能否滿流
程式碼:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 3010;
const int maxm = 50010;
const int inf = 0x3f3f3f3f;
int p,r,c;
int D[maxn];
int E[maxn];
int I[maxm];
int J[maxm];
int T[maxm];
int tol,head[maxn];
struct edge
{
int to,next,cap,flow;
}es[maxm];
void addedge( int u , int v , int w )
{
es[tol].to = v;
es[tol].cap = w;
es[tol].flow = 0;
es[tol].next = head[u];
head[u] = tol++;
es[tol].to = u;
es[tol].cap = 0;
es[tol].flow = 0;
es[tol].next = head[v];
head[v] = tol++;
}
int gap[maxn],dep[maxn],cur[maxn],Q[maxn],S[maxn];
void bfs( int s , int t )
{
memset ( dep , -1 , sizeof(dep) );
memset ( gap , 0 , sizeof(gap) );
gap[0] = 1;
int front = 0,rear = 0;
dep[t] = 0;
Q[rear++] = t;
while ( front!=rear )
{
int u = Q[front++];
for( int i=head[u] ; i!=-1 ; i=es[i].next )
{
int v = es[i].to;
if ( dep[v]!=-1 ) continue;
Q[rear++] = v;
dep[v] = dep[u]+1;
gap[dep[v]]++;
}
}
}
int sap( int s , int t , int n )
{
bfs( s , t );
memcpy( cur , head , sizeof(cur) );
int top = 0,u = s,ans = 0;
while( dep[s]<n )
{
if ( u==t )
{
int minx = inf;
int inser;
for( int i=0 ; i<top ; i++ )
if ( minx>es[S[i]].cap-es[S[i]].flow )
{
minx = es[S[i]].cap-es[S[i]].flow;
inser = i;
}
for( int i=0 ; i<top ; i++ )
{
es[S[i]].flow += minx;
es[S[i]^1].flow -= minx;
}
ans += minx;
top = inser;
u = es[S[top]^1].to;
continue;
}
bool flag = false;
int v;
for ( int i=cur[u] ; i!=-1 ; i=es[i].next )
{
v = es[i].to;
if ( es[i].cap-es[i].flow&&dep[v]+1==dep[u] )
{
flag = true;
cur[u] = i;
break;
}
}
if ( flag )
{
S[top++] = cur[u];
u = v;
continue;
}
int Min = n;
for ( int i=head[u] ; i!=-1 ; i=es[i].next )
if ( es[i].cap-es[i].flow&&dep[es[i].to]<Min )
{
Min = dep[es[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if ( !gap[dep[u]] ) return ans;
dep[u] = Min+1;
gap[dep[u]]++;
if ( u!=s ) u = es[S[--top]^1].to;
}
return ans;
}
bool ok( int limit , int sumx )
{
tol = 0; memset( head , -1 , sizeof(head) );
int s = 0,t = p+r+1;
for ( int i=1 ; i<=p ; i++ )
addedge( s , i , D[i] );
for ( int i=1 ; i<=r ; i++ )
addedge( i+p , t , E[i] );
for ( int i=1 ; i<=c ; i++ )
if ( T[i]<=limit ) addedge( I[i] , J[i]+p , inf );
return sap( s , t , t+1 )==sumx;
}
int main()
{
for ( ; scanf( "%d%d%d" , &p , &r , &c )==3 ; )
{
int sumd = 0,sume = 0;
for ( int i=1 ; i<=p ; i++ ) scanf( "%d" , &D[i] ),sumd += D[i];
for ( int i=1 ; i<=r ; i++ ) scanf( "%d" , &E[i] ),sume += E[i];
for ( int i=1 ; i<=c ; i++ ) scanf( "%d%d%d" , &I[i] , &J[i] , &T[i] );
int l=1,r=1000000;
while ( l<=r )
{
int mid = (l+r)>>1;
if ( ok(mid,sumd) ) r = mid-1;
else l = mid+1;
}
if ( l>1000000 ) printf( "-1\n" );
else printf( "%d\n" , l );
}
return 0;
}
題意:有n個關燈序列和m盞燈,按關燈序列開關燈,問最少個關燈序列可以使得所有點全滅
思路:關燈序列不可能遍歷2*n次以上,暴力開關燈
程式碼:
#include<bits/stdc++.h>
using namespace std;
bool on[1010];
int a[1010][1010];
int main()
{
int n,m;
scanf ( "%d%d" , &n , &m );
int ans = -1;
for ( int i=1 ; i<=m ; i++ )
on[i] = false;
int s;
scanf( "%d" , &s );
for ( int i=1 ; i<=s ; i++ )
{
int x;
scanf( "%d" , &x );
on[x] = true;
}
for ( int i=1 ; i<=n ; i++ )
{
scanf( "%d" , &a[i][0] );
for ( int j=1 ; j<=a[i][0] ; j++ )
scanf ( "%d" , &a[i][j] );
}
for ( int i=1 ; i<=n+n ; i++ )
{
int t = i;
if ( t>n ) t = t-n;
for( int j=1 ; j<=a[t][0] ; j++ )
on[a[t][j]] = !on[a[t][j]];
bool ok = true;
for( int j=1 ; j<=m ; j++ )
if ( on[j] ) ok = false;
if ( ok&&ans==-1 ) ans = i;
}
printf( "%d\n" , ans );
return 0;
}
題意:詢問樹上兩條路徑的公共節點數
思路:若兩條路徑有公共節點則其中一條路徑的最近公共祖先一定在另一條路徑上。兩條路徑的交匯點一定是從四個點的公共祖先之間。
程式碼:
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
typedef pair<int,int> pii;
const int maxn = 100010;
const int maxm = 200010;
int tol,head[maxn];
struct edge
{
int to,next;
}es[maxm];
void addedge( int u , int v )
{
es[tol].to = v;
es[tol].next = head[u];
head[u] = tol++;
}
int fa[maxn][20],dep[maxn];
void dfs( int u , int f , int d )
{
for ( int i=head[u] ; i!=-1 ; i=es[i].next )
{
int v = es[i].to; if ( v==f ) continue;
fa[v][0] = u; dep[v] = d+1; dfs( v , u , d+1 );
}
}
int lca( int u , int v )
{
if ( dep[u]<dep[v] ) swap ( u , v );
int d = dep[u]-dep[v];
for ( int i=19 ; i>=0 ; i-- )
if ( d&(1<<i) ) u = fa[u][i];
if ( u==v ) return u;
for ( int i=19 ; i>=0 ; i-- )
if ( fa[u][i]!=fa[v][i] )
{
u = fa[u][i];
v = fa[v][i];
}
return fa[u][0];
}
int main()
{
tol = 0; memset( head , -1 , sizeof(head) );
int n; scanf( "%d" , &n );
int q; scanf( "%d" , &q );
for ( int i=2 ; i<=n ; i++ )
{
int u,v; scanf( "%d%d" , &u , &v );
addedge( u , v );
addedge( v , u );
}
fa[1][0] = 0; dep[1] = 1; dfs( 1 , 0 , 1 );
for ( int j=1 ; j<20 ; j++ )
for ( int i=1 ; i<=n ; i++ )
fa[i][j] = fa[fa[i][j-1]][j-1];
for ( int i=0 ; i< q ; i++ )
{
int a,b; scanf( "%d%d" , &a , &b );
int c,d; scanf( "%d%d" , &c , &d );
int ans = 0; bool ok = false;
int f1 = lca( a , b );
int f2 = lca( c , d );
if ( dep[f1]<dep[f2] ) swap( f1 , f2 );
int ff1 = lca( a , c );
int ff2 = lca( a , d );
int ff3 = lca( b , c );
int ff4 = lca( b , d );
if ( dep[ff1]>=dep[f1] )
{
ok = true; ans -= dep[f1]-dep[ff1];
}
if ( dep[ff2]>=dep[f1] )
{
ok = true; ans -= dep[f1]-dep[ff2];
}
if ( dep[ff3]>=dep[f1] )
{
ok = true; ans -= dep[f1]-dep[ff3];
}
if ( dep[ff4]>=dep[f1] )
{
ok = true; ans -= dep[f1]-dep[ff4];
}
if ( ok ) printf( "%d\n" , ans+1 );
else printf( "%d\n" , ans );
}
return 0;
}