T1：糖果机器

solution

考虑从第$i$个糖果出发能到达第$j$个，则有$T_j-T_i\ge |S_j-S_i|$
$$\{\begin{array}{r}{T}_j-T_i\ge S_j-S_i\\ T_j-T_i\ge S_i-S_j\end{array}=>\{\begin{array}{r}{T}_i+S_i\ge T_j+S_j\\ T_i-S_i\ge T_j-S_j\end{array}$$
发现这其实是函数同构体

令$A=T_i+S_i,B=T_i-S_i$，则👆可转化为👇
$$\{\begin{array}{r}{A}_i\ge A_j\\ B_i\ge B_j\end{array}$$
又发现此题就是一个二维偏序问题
其实就是noip普及组1999导弹拦截

二维偏序问题，一般是一维时间轴有序，然后二维套树状数组，总体上复杂度是$log$级别

code

#include <set>
#include <cstdio>
#include <algorithm>
using namespace std;
#define maxn 100005
struct node {int s, t, id;friend bool operator < ( const node &x, const node &y ) {return x.s + x.t < y.s + y.t;}
}p[maxn];
set < node > st;
int n, tot;bool cmp( node x, node y ) {return ( x.t - x.s == y.t - y.s ) ? ( x.t + x.s < y.t + y.s ) : ( x.t - x.s < y.t - y.s );
}int main() {scanf( "%d", &n );for( int i = 1;i <= n;i ++ )scanf( "%d %d", &p[i].s, &p[i].t );sort( p + 1, p + n + 1, cmp );for( int i = 1;i <= n;i ++ ) {set < node > :: iterator it = st.upper_bound( p[i] );if( it == st.begin() ) p[i].id = ++ tot;else it --, p[i].id = it->id, st.erase( it );st.insert( p[i] );}printf( "%d\n", tot );for( int i = 1;i <= n;i ++ )printf( "%d %d %d\n", p[i].s, p[i].t, p[i].id );return 0;
}

T2：手套

solution

code

#include <cstdio>
#include <algorithm>
using namespace std;
#define maxn 20
struct node {int x, y;
}p[1 << maxn];
int n;
int a[maxn], b[maxn];bool cmp( node s, node t ) {return ( s.x == t.x ) ? ( s.y > t.y ) : ( s.x < t.x );
}void update( long long &ansx, long long &ansy, long long x, long long y ) {if( ( x + y < ansx + ansy ) || ( x + y == ansx + ansy && x < ansx ) )ansx = x, ansy = y;
}int main() {scanf( "%d", &n );for( int i = 0;i < n;i ++ )scanf( "%d", &a[i] );for( int i = 0;i < n;i ++ )scanf( "%d", &b[i] );int m = 1 << n;for( int i = 0;i < m;i ++ )for( int j = 0;j < n;j ++ )if( i >> j & 1 ) p[i].x += a[j];for( int i = 0;i < m;i ++ )for( int j = 0;j < n;j ++ )if( i >> j & 1 ) p[m - 1 - i].y += b[j];long long flagx = p[m - 1].x, flagy = p[0].y, ansx = flagx, ansy = flagy;sort( p, p + m, cmp );for( int i = m - 1, y = 0;~ i;i -- ) {if( p[i].x != flagx && y != flagy )update( ansx, ansy, p[i].x + 1, y + 1 );y = max( p[i].y, y );}printf( "%lld\n%lld\n", ansx, ansy );return 0;
}

T3：甲虫

solution

区间dp还是一眼就看得出来，只不过不知道怎么处理时间问题，考场上错误代码都能骗40

正着不行，就反着考虑

code

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 305
#define ll long long
#define inf 0x3f3f3f3f
int n, m, k;
int x[maxn];
ll dp[2][maxn][maxn];
int vis[2][maxn][maxn];long long dfs( int t, int l, int r ) {if( r - l == k ) return 0;if( vis[t][l][r] == k ) return dp[t][l][r];vis[t][l][r] = k;int p = ( t ? r : l );ll &ans = dp[t][l][r];ans = inf;if( l != 1 ) ans = min( ans, dfs( 0, l - 1, r ) + ( x[p] - x[l - 1] ) * ( k - ( r - l ) ) );if( r != n ) ans = min( ans, dfs( 1, l, r + 1 ) + ( x[r + 1] - x[p] ) * ( k - ( r - l ) ) );return ans;
}int main() {scanf( "%d %d", &n, &m );for( int i = 1;i <= n;i ++ )scanf( "%d", &x[i] );x[++ n] = 0;sort( x + 1, x + n + 1 );int pos;for( int i = 1;i <= n;i ++ )if( ! x[i] ) { pos = i; break; }ll ans = 0;for( k = 0;k < n;k ++ )ans = max( ans, 1ll * m * k - dfs( 0, pos, pos ) );printf( "%lld", ans );return 0;
}

T4：选举

solution

code

#include <cstdio>
#include <iostream>
using namespace std;
#define maxn 500005
struct node {int minn, maxx, ans;node() {}node( int Min, int Max, int Ans ) {minn = Min, maxx = Max, ans = Ans;}friend node operator + ( const node &x, const node &y ) {return node( min( x.minn, y.minn ), max( x.maxx, y.maxx ), max( y.maxx - x.minn, max( x.ans, y.ans ) ) );}
}t[maxn << 2];
int n, Q;
char s[maxn];
int pre[maxn];void build( int num, int l, int r ) {if( l == r ) {t[num] = node( pre[l], pre[l], 0 );return;}int mid = ( l + r ) >> 1;build( num << 1, l, mid ), build( num << 1 | 1, mid + 1, r );t[num] = t[num << 1] + t[num << 1 | 1];
}node query( int num, int l, int r, int L, int R ) {if( L <= l && r <= R ) return t[num];int mid = ( l + r ) >> 1;if( R <= mid ) return query( num << 1, l, mid, L, R );else if( mid < L ) return query( num << 1 | 1, mid + 1, r, L, R );else return query( num << 1, l, mid, L, R ) + query( num << 1 | 1, mid + 1, r, L, R );
}int main() {scanf( "%d %s %d", &n, s + 1, &Q );for( int i = 1;i <= n;i ++ )pre[i] = pre[i - 1] + ( s[i] == 'T' ? -1 : 1 );build( 1, 0, n );	while( Q -- ) {int l, r;scanf( "%d %d", &l, &r );node ans = query( 1, 0, n, l - 1, r );printf( "%d\n", ans.ans - ( pre[r] - pre[l - 1] ) );}return 0;
}