Educational Codeforces Round 60 Div. 2
F：考虑对于每个字母对求出删掉哪些字符集会造成字符串不合法，只要考虑相邻出现的该字母对即可，显然这可以在O(np2)（或小常数O(np3)）内求出。然后再对每个字符集判断是否能通过一步删除转移而来即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
#define M 17
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,c[N],b[N],cnt[M],ans;
char s[N];
bool a[M][M],f[1<<M],g[1<<M];
signed main()
{
#ifndef ONLINE_JUDGE
freopen("f.in","r",stdin);
freopen("f.out","w",stdout);
const char LL[]="%I64d\n";
#endif
n=read(),m=read();
scanf("%s",s+1);for (int i=1;i<=n;i++) cnt[c[i]=s[i]-'a']++;
for (int i=0;i<m;i++)
for (int j=0;j<m;j++)
{
a[i][j]=read();
if (i<=j&&!a[i][j])
{
memset(g,0,sizeof(g));int last=-1,s=0;
for (int x=1;x<=n;x++)
{
if (c[x]==i||c[x]==j)
{
if (last==i+j-c[x]) g[s]=1;
last=c[x];s=0;
}
else s|=(1<<c[x]);
}
for (int x=1;x<(1<<m);x++)
if (!g[x]&&!(x&(1<<i))&&!(x&(1<<j)))
{
for (int t=x,y=t&-t;t;t^=y,y=t&-t)
g[x]|=g[x^y];
}
for (int x=0;x<(1<<m);x++) f[x]|=g[x];
}
}
ans=n;
for (int i=1;i<(1<<m);i++)
{
if (!f[i])
{
f[i]=1;
for (int j=0;j<m;j++)
if ((i&(1<<j))&&!f[i^(1<<j)]) {f[i]=0;break;}
}
if (!f[i])
{
int tot=n;
for (int j=0;j<m;j++) if (i&(1<<j)) tot-=cnt[j];
ans=min(ans,tot);
}
}
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
G：考虑一个数被删掉对一个区间产生的贡献，只考虑某一侧，显然如果该区间内该数为最大值，贡献即为该侧区间长度，否则为其到下一个比他大的数的距离。离线后扫过去维护贡献，线段树实现区间加一次函数即可。
又卡我常！
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1000010
#define tree lazy
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,a[N],nxt[N],L[N<<2],R[N<<2];
ll ans[N],lazy[N<<2][2];
struct data{int l,r,i;ll ans;
}q[N];
bool cmp(const data&a,const data&b)
{
return a.l<b.l;
}
void build(int k,int l,int r)
{
L[k]=l,R[k]=r;lazy[k][0]=lazy[k][1]=0;
if (l==r) return;
int mid=l+r>>1;
build(k<<1,l,mid);
build(k<<1|1,mid+1,r);
}
void down(int k)
{
tree[k<<1][0]+=tree[k][0],tree[k<<1|1][0]+=tree[k][0],tree[k][0]=0;
tree[k<<1][1]+=tree[k][1],tree[k<<1|1][1]+=tree[k][1],tree[k][1]=0;
}
ll query(int k,int p)
{
if (L[k]==R[k]) return 1ll*lazy[k][1]*p+lazy[k][0];
down(k);
int mid=L[k]+R[k]>>1;
if (p<=mid) return query(k<<1,p);
else return query(k<<1|1,p);
}
void add(int k,int l,int r,int u,int x)
{
if (L[k]==l&&R[k]==r) {lazy[k][1]++;lazy[k][0]+=x;return;}
down(k);
int mid=L[k]+R[k]>>1;
if (r<=mid) add(k<<1,l,r,u,x);
else if (l>mid) add(k<<1|1,l,r,u,x);
else add(k<<1,l,mid,u,x),add(k<<1|1,mid+1,r,u,x);
}
void add(int k,int l,int r,int x)
{
if (L[k]==l&&R[k]==r) {lazy[k][0]+=x;return;}
down(k);
int mid=L[k]+R[k]>>1;
if (r<=mid) add(k<<1,l,r,x);
else if (l>mid) add(k<<1|1,l,r,x);
else add(k<<1,l,mid,x),add(k<<1|1,mid+1,r,x);
}
void solve()
{
nxt[n+1]=n+1;
for (int i=n;i>=1;i--)
{
int j=i+1;
while (a[j]<a[i]) j=nxt[j];
nxt[i]=j;
}
build(1,1,n);
sort(q+1,q+m+1,cmp);
int cur=0;
for (int i=1;i<=n;i++)
{
while (q[cur+1].l==i) cur++,q[cur].ans-=query(1,q[cur].r);
add(1,i,nxt[i]-1,1,-i);
if (nxt[i]<=n) add(1,nxt[i],n,nxt[i]-i-1);
}
for (int i=1;i<=m;i++) q[i].ans+=query(1,q[i].r);
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("g.in","r",stdin);
freopen("g.out","w",stdout);
#endif
n=read(),m=read();
for (int i=1;i<=n;i++) a[i]=read();
a[0]=a[n+1]=n+1;
for (int i=1;i<=m;i++) q[i].l=read();
for (int i=1;i<=m;i++) q[i].r=read();
for (int i=1;i<=m;i++) q[i].i=i;
solve();
reverse(a+1,a+n+1);for (int i=1;i<=m;i++) q[i].l=n-q[i].l+1,q[i].r=n-q[i].r+1,swap(q[i].l,q[i].r);
solve();
for (int i=1;i<=m;i++) q[i].ans+=q[i].r-q[i].l+1;
for (int i=1;i<=m;i++) ans[q[i].i]=q[i].ans;
for (int i=1;i<=m;i++) printf("%I64d ",ans[i]);
return 0;
//NOTICE LONG LONG!!!!!
}
场上什么傻逼题都写不出来，自闭了。
