计算组合数然后多重集排列乱搞，注意判无解情况（TM我就判错然后FST了）。
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
using namespace std;
inline int read() {
int x=0,f=1;char c=getchar();
for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
for(;isdigit(c);c=getchar()) x=x*10+c-'0';
return x*f;
}
typedef long long ll;
const int maxn=1010;
const int mod=1000000007;
int n,c[26];
char s[maxn];
ll C[maxn][maxn];
void solve() {
memset(c,0,sizeof(c));
scanf("%s",s+1);n=strlen(s+1);
rep(i,1,n) c[s[i]-'a']++;
int sum=0;
rep(i,0,25) {
if(c[i]&1) sum++,c[i]--;
c[i]>>=1;
}
if(sum>1) {puts("0");return;}
ll ans=1;int cur=0;
rep(i,0,25) {
cur+=c[i];
(ans*=C[cur][c[i]])%=mod;
}
printf("%I64d\n",ans);
}
int main() {
C[0][0]=1;
rep(i,1,1000) {
C[i][0]=C[i][i]=1;
rep(j,1,i-1) C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
}
dwn(T,read(),1) solve();
return 0;
}
转化成对偶图然后统计八连块联通性。
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
using namespace std;
inline int read() {
int x=0,f=1;char c=getchar();
for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
for(;isdigit(c);c=getchar()) x=x*10+c-'0';
return x*f;
}
typedef long long ll;
const int mx[]={0,0,1,-1,1,-1,1,-1};
const int my[]={1,-1,0,0,1,1,-1,-1};
const int maxn=510;
int pa[maxn*maxn],n,m;
int id(int i,int j) {return (i-1)*m+j;}
char s[maxn];
int findset(int x) {return pa[x]==x?x:pa[x]=findset(pa[x]);}
void Add(int x,int y) {
pa[id(x,y)]=id(x,y);
rep(dir,0,7) {
int nx=x+mx[dir],ny=y+my[dir];
if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&pa[id(nx,ny)]) pa[findset(id(nx,ny))]=findset(id(x,y));
}
}
void solve() {
n=read();m=read();int S=n*m+1,T=n*m+2;
memset(pa,0,sizeof(pa));pa[S]=S;pa[T]=T;
rep(i,1,n) {
scanf("%s",s+1);
rep(j,1,m) {
if(s[j]=='1') Add(i,j);
if(j==1&&s[j]=='1') pa[findset(S)]=findset(id(i,j));
if(j==m&&s[j]=='1') pa[findset(T)]=findset(id(i,j));
}
}
int k=read(),ok=0;
if(findset(S)==findset(T)) {puts("0");ok=1;}
rep(i,1,k) {
int x=read()+1,y=read()+1;
Add(x,y);
if(y==1) pa[findset(S)]=findset(id(x,y));
if(y==m) pa[findset(T)]=findset(id(x,y));
if(!ok&&findset(S)==findset(T)) {printf("%d\n",i);ok=1;}
}
if(!ok) printf("-1\n");
}
int main() {
dwn(T,read(),1) solve();
return 0;
}
T3 Bomber Man wants to bomb an Array.
直接DP，发现一个转移是否合法只取决于区间内是否恰好有一个点。
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#include<algorithm>
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
using namespace std;
inline int read() {
int x=0,f=1;char c=getchar();
for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
for(;isdigit(c);c=getchar()) x=x*10+c-'0';
return x*f;
}
typedef long long ll;
const int maxn=2010;
double score(int len) {return log2(len);}
int n,m,S[maxn];
double f[maxn];
void solve() {
n=read();m=read();
memset(S,0,sizeof(S));
rep(i,1,m) S[read()+1]++;
rep(i,1,n) S[i]+=S[i-1];
f[0]=0.0;
rep(i,1,n) {
f[i]=0.0;
rep(j,0,i-1) if(S[i]-S[j]==1) f[i]=max(f[i],f[j]+score(i-j));
}
printf("%.0lf\n",floor(1000000*f[n]));
}
int main() {
dwn(T,read(),1) solve();
return 0;
}
T4 xiaoxin and his watermelon candy
先暴力哈希，然后转化成区间不同数的个数查询，离线树状数组随便做做。
#include<cstdio>
#include<cstring>
#include<cctype>
#include<ctime>
#include<algorithm>
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
using namespace std;
const int BufferSize=1<<16;
char buffer[BufferSize],*head,*tail;
inline char Getchar() {
if(head==tail) {
int l=fread(buffer,1,BufferSize,stdin);
tail=(head=buffer)+l;
}
return *head++;
}
inline int read() {
int x=0,f=1;char c=Getchar();
for(;!isdigit(c);c=Getchar()) if(c=='-') f=-1;
for(;isdigit(c);c=Getchar()) x=x*10+c-'0';
return x*f;
}
typedef long long ll;
typedef unsigned long long ull;
const int maxn=200010;
ull X,val[maxn],tmp[maxn];
int n,m,A[maxn];
struct Query {
int l,r,id;
bool operator < (const Query& ths) const {
return l<ths.l;
}
}Q[maxn];
int last[maxn],f[maxn];
int sumv[maxn],ans[maxn];
int sum(int x) {int res=0;for(;x;x-=x&-x) res+=sumv[x];return res;}
void add(int x,int v) {for(;x<=n;x+=x&-x) sumv[x]+=v;}
void solve() {
n=read();
rep(i,1,n) A[i]=read()+1,sumv[i]=0;
if(n>=3) {
rep(i,1,n-2) {
if(A[i]<=A[i+1]&&A[i+1]<=A[i+2]) val[i]=X*X*A[i]+X*A[i+1]+(ull)A[i+2];
else val[i]=0;
tmp[i]=val[i];
}
sort(tmp+1,tmp+n-1);
rep(i,1,n-2) {
if(!val[i]) A[i]=0;
else A[i]=lower_bound(tmp+1,tmp+n-1,val[i])-tmp;
}
rep(i,0,n) last[i]=n+1;
dwn(i,n-2,1) f[i]=last[A[i]],last[A[i]]=i;
rep(i,1,n) if(last[i]<n) add(last[i],1);
m=read();
rep(i,1,m) Q[Q[i].id=i].l=read(),Q[i].r=read()-2;
sort(Q+1,Q+m+1);int cur=1;
rep(i,1,m) {
while(cur<Q[i].l&&cur<=n-2) {
if(A[cur]) add(cur,-1),add(f[cur],1);
cur++;
}
if(Q[i].l<=Q[i].r) ans[Q[i].id]=sum(Q[i].r)-sum(Q[i].l-1);
else ans[Q[i].id]=0;
}
rep(i,1,m) printf("%d\n",ans[i]);
}
else {
m=read();
rep(i,1,m) read(),read(),printf("%d\n",0);
}
}
int main() {
X=52501;srand(time(0));
dwn(i,rand()%10+5,1) X=X*233333;
dwn(T,read(),1) solve();
return 0;
}