三元组
solution
60分
\(s[i]\)表示以\(i\)结尾的回文串的首项下标和.
\(t[i]\)表示以\(i\)开头的回文串的末项下标和.
答案为\(sum_{i=1}^{|S|-1}s[i]\times{t[i+1]}\).
\(manacher\)求出以每个点为中心的最长回文串半径,此回文串范围内对\(s[\;],t[\;]\)区间覆盖等差数列.
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 2000005
#define M 1000000007
using namespace std;
int r[N<<1],s[N],t[N],n,ti,ans;
char a[N<<1];
inline void add_s(int l,int r,int m){
for(int i=l;i<=r;++i)
s[i]+=m-i;
}
inline void add_t(int l,int r,int m){
for(int i=l;i<=r;++i)
t[i]+=m-i;
}
inline void manacher(){
int mx=0,id=0;
m=n=strlen(a+1);
for(int i=n;i;--i){
a[i<<1]=a[i];a[i<<1|1]='#';
}
n=n<<1|1;
a[0]='$';a[1]='#';
for(int i=1;i<=n;++i){
r[i]=i<mx?min(r[(id<<1)-i],mx-i):1;
while(a[i+r[i]]==a[i-r[i]]) ++r[i];
if(i+r[i]>mx) mx=i+r[i],id=i;
add_s(i+1>>1,(i>>1)+(r[i]-1>>1),i);
add_t((i+1>>1)-(r[i]-1>>1),i>>1,i);
}
}
inline void Aireen(){
scanf("%d",&ti);
while(ti--){
scanf("%s",a+1);
n=strlen(a+1);
memset(s,0,sizeof(s));
memset(t,0,sizeof(t));
manacher();
ans=0ll;
for(int i=2;i<=n;++i){
ans+=1ll*s[i-1]*t[i]%M;
if(ans>M) ans-=M;
}
printf("%d\n",ans);
}
}
int main(){
freopen("triple.in","r",stdin);
freopen("triple.out","w",stdout);
Aireen();
fclose(stdin);
fclose(stdout);
return 0;
}
100分
利用差分的思想\(O(1)\)覆盖等差数列,统计的时候计算前缀和即可.
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 1000005
#define M 1000000007
using namespace std;
int r[N<<1],s[N],t[N],ks[N],kt[N],n,m,ti,ans;
char a[N<<1];
inline void add_s(int l,int r,int m){
if(l>r) return;
s[l]+=m-l;s[r+1]-=m-r;
if(s[l]>M) s[l]-=M;
if(s[r+1]<-M) s[r+1]+=M;
--ks[l+1];++ks[r+1];
}
inline void add_t(int l,int r,int m){
if(l>r) return;
t[l]+=m-l;t[r+1]-=m-r;
if(t[l]>M) t[l]-=M;
if(t[r+1]<-M) t[r+1]+=M;
--kt[l+1];++kt[r+1];
}
inline void manacher(){
int mx=0,id=0;
m=n=strlen(a+1);
for(int i=n;i;--i){
a[i<<1]=a[i];a[i<<1|1]='#';
}
n=n<<1|1;
a[0]='$';a[1]='#';
for(int i=1;i<=n;++i){
r[i]=i<mx?min(r[(id<<1)-i],mx-i):1;
while(a[i+r[i]]==a[i-r[i]]) ++r[i];
if(i+r[i]>mx) mx=i+r[i],id=i;
add_s(i+1>>1,(i>>1)+(r[i]-1>>1),i);
add_t((i+1>>1)-(r[i]-1>>1),i>>1,i);
}
}
inline void Aireen(){
scanf("%d",&ti);
while(ti--){
scanf("%s",a+1);
memset(s,0,sizeof(s));
memset(t,0,sizeof(t));
memset(ks,0,sizeof(ks));
memset(kt,0,sizeof(kt));
manacher();ans=0ll;
for(int i=1,k1=0,k2=0;i<=m;++i){
k1+=ks[i];
s[i]+=s[i-1]+k1;
if(s[i]>M) s[i]-=M;
k2+=kt[i];
t[i]+=t[i-1]+k2;
if(t[i]>M) t[i]-=M;
}
for(int i=2;i<=m;++i){
ans+=1ll*s[i-1]*t[i]%M;
if(ans>M) ans-=M;
}
printf("%d\n",ans);
}
}
int main(){
freopen("triple.in","r",stdin);
freopen("triple.out","w",stdout);
Aireen();
fclose(stdin);
fclose(stdout);
return 0;
}
最优价值
solution
最大权闭合子图
对于边\((u,v)\),如果选择\(u\),必须选择\(v\).
建图
对于原图中的边\((u,v)\),连边\((u,v)=+\infty\).
如果点\(u\)权\(w\)为正,连边\((s,u)=w\);否则连边\((u,t)=-w\).
\(w_{max}=\sum_{w_i>0}{w_i}-Mincut\)
100分
分为三类点:\((i,j)=w(i,j),i=-a,x=-b_x+a_x\).
\((i,j)\)连向\(i,j;i\)连向\(s[i]\).
用最大权闭合子图做即可.
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define K 110
#define N 100000
#define M 1000000
#define INF 1000000000
#define min(a,b) (a<b?a:b)
using namespace std;
struct graph{
int nxt,to,f;
}e[M];
int w[K][K],p[K][K],c[K],a[10],b[10],g[N],dep[N],n,m,k,s,t,cnt,sum;
char ch[K];
queue<int> q;
inline void addedge(int x,int y,int f){
e[++cnt].nxt=g[x];g[x]=cnt;e[cnt].to=y;e[cnt].f=f;
}
inline void adde(int x,int y,int f){
addedge(x,y,f);addedge(y,x,0);
}
inline bool bfs(int u){
memset(dep,0,sizeof(dep));
dep[u]=1;q.push(u);
while(!q.empty()){
u=q.front();q.pop();
for(int i=g[u];i;i=e[i].nxt)
if(e[i].f>0&&!dep[e[i].to]){
dep[e[i].to]=dep[u]+1;q.push(e[i].to);
}
}
return dep[t];
}
inline int dfs(int u,int f){
if(u==t) return f;
int ret=0;
for(int i=g[u],d;i&&f;i=e[i].nxt)
if(e[i].f>0&&dep[e[i].to]>dep[u]){
d=dfs(e[i].to,min(f,e[i].f));
f-=d;ret+=d;e[i].f-=d;e[i^1].f+=d;
}
if(!ret) dep[u]=-1;
return ret;
}
inline int dinic(){
int ret=0;
while(bfs(s)) ret+=dfs(s,INF);
return ret;
}
inline void Aireen(){
scanf("%d",&m);
while(m--){
scanf("%d",&n);
k=n*(n-1)/2;
s=k+n+11;t=s+1;
scanf("%s",ch+1);
for(int i=1;i<=n;++i)
c[i]=ch[i]-'0';
for(int i=0;i<=9;++i)
scanf("%d%d",&a[i],&b[i]);
sum=0;
for(int i=1;i<=n;++i)
for(int j=1;j<=n;++j)
scanf("%d",&w[i][j]);
cnt=0;
for(int i=1;i<n;++i)
for(int j=i+1;j<=n;++j)
p[i][j]=++cnt;
cnt=1;
memset(g,0,sizeof(g));
for(int i=1;i<n;++i)
for(int j=i+1;j<=n;++j){
sum+=w[i][j]+w[j][i];
adde(s,p[i][j],w[i][j]+w[j][i]);
adde(p[i][j],k+i,INF);adde(p[i][j],k+j,INF);
}
for(int i=1;i<=n;++i){
adde(k+i,k+n+c[i]+1,INF);
adde(k+i,t,a[c[i]]);
}
for(int i=0;i<=9;++i)
adde(k+n+i+1,t,b[i]-a[i]);
printf("%d\n",sum-dinic());
}
}
int main(){
freopen("value.in","r",stdin);
freopen("value.out","w",stdout);
Aireen();
fclose(stdin);
fclose(stdout);
return 0;
}