I'm good food ah. . . . .
92 will only fight violence, violence will not be a 93
Analog 92,
T1: direct ex_gcd plus Category talk to
T2: the examination room only storm hit search, after being interpreted as sort segment tree to solve, sort keywords for a + b, because if ai <bj && bi <aj so should i before j.
1 #include<bits/stdc++.h> 2 #define N 100050 3 #define LL long long 4 using namespace std; 5 int n,pd[N],cnt,lsh[N<<1],ls; 6 long long dp[N],ans; 7 struct node{ 8 int a,b,w; 9 friend bool operator <(const node &x,const node &y) 10 { 11 return x.a+x.b<y.a+y.b; 12 } 13 }q[N]; 14 inline void init() 15 { 16 for(int i=1;i<=n;++i)lsh[++ls]=q[i].a,lsh[++ls]=q[i].b; 17 sort(lsh+1,lsh+ls+1);ls=unique(lsh+1,lsh+ls+1)-lsh-1; 18 for(int i=1;i<=n;++i){ 19 q[i].a=lower_bound(lsh+1,lsh+ls+1,q[i].a)-lsh; 20 q[i].b=lower_bound(lsh+1,lsh+ls+1,q[i].b)-lsh; 21 }sort(q+1,q+n+1); 22 } 23 LL ma[N<<3],tag[N<<3]; 24 inline void plu(int g,LL w){ma[g]+=w,tag[g]+=w;} 25 inline void upd(int g){ma[g]=max(ma[g<<1],ma[g<<1|1]);} 26 inline void down(int g){plu(g<<1,tag[g]),plu(g<<1|1,tag[g]);tag[g]=0;} 27 void add(int g,int l,int r,int x,int y,int w) 28 { 29 if(l>y||r<x)return;if(l>=x&&r<=y)return plu(g,w); 30 if(tag[g])down(g);const int m=l+r>>1; 31 add(g<<1,l,m,x,y,w);add(g<<1|1,m+1,r,x,y,w); 32 upd(g); 33 } 34 void change(int g,int l,int r,int pos,LL w) 35 { 36 if(l==r)return (void)(ma[g]=max(ma[g],w)); 37 if(tag[g])down(g);const int m=l+r>>1; 38 if(pos<=m)change(g<<1,l,m,pos,w); 39 else change(g<<1|1,m+1,r,pos,w); 40 upd(g); 41 } 42 LL ask(int g,int l,int r,int x,int y) 43 { 44 if(l>y||r<x)return 0; 45 if(l>=x&&r<=y)return ma[g]; 46 if(tag[g])down(g); 47 const int m=l+r>>1; 48 const LL a1=ask(g<<1,l,m,x,y),a2=ask(g<<1|1,m+1,r,x,y); 49 return max(a1,a2); 50 } 51 inline void duizhangkuaipao() 52 { 53 for(int i=1;i<=n;++i) 54 { 55 dp[i]=ask(1,1,ls,1,min(q[i].a,q[i].b))+q[i].w; 56 add(1,1,ls,q[i].a,q[i].b,q[i].w); 57 change(1,1,ls,q[i].a,dp[i]); 58 } 59 ans=ma[1]; 60 } 61 int main() 62 { 63 scanf("%d",&n); 64 for(int i=1;i<=n;++i) 65 scanf("%d%d%d",&q[i].a,&q[i].b,&q[i].w); 66 init(); 67 duizhangkuaipao(); 68 cout<<ans<<endl; 69 }
T3: Multi-source shortest, for each point to maintain this particular point and the corresponding point of the shortest distance is that, when the edge scan update the answers to each
1 #include<bits/stdc++.h> 2 #define N 200050 3 #define LL long long 4 using namespace std; 5 int n,m,p,a[N],pd[N],bl[N]; 6 const LL inf=10000000000000000; 7 LL dis[N],ans[N]; 8 int he[N],ne[N<<2],to[N<<2],w[N<<2],tot; 9 inline void work1() 10 { 11 for(int i=1,x,y,z;i<=m;++i){ 12 scanf("%d%d%d",&x,&y,&z); 13 ans[x]=min(ans[x],(LL)z); 14 ans[y]=min(ans[y],(LL)z); 15 } 16 for(int i=1;i<=p;++i)printf("%lld ",ans[a[i]]); 17 } 18 inline void addedge(int x,int y,int z) 19 { 20 to[++tot]=y;ne[tot]=he[x]; 21 w[tot]=z;he[x]=tot; 22 } 23 priority_queue<pair<LL,int> >q; 24 #define mmp make_pair 25 #define fir first 26 #define sec second 27 inline void getans() 28 { 29 for(int i=0;i<=n;++i)dis[i]=inf; 30 for(int i=1;i<=p;++i) 31 dis[a[i]]=0,bl[a[i]]=a[i],q.push(mmp(0,a[i])); 32 LL d;int g; 33 while(q.size()) 34 { 35 g=q.top().sec; 36 d=-q.top().fir; 37 q.pop(); 38 if(dis[g]!=d)continue; 39 // printf("g:%d d:%lld\n",g,d); 40 for(int i=he[g];i;i=ne[i]){ 41 if(bl[g]&&bl[to[i]]&&bl[to[i]]!=bl[g]){ 42 // printf("g:%d to:%d blg:%d blt:%d new:%lld\n",g,to[i],bl[g],bl[to[i]],d+w[i]+dis[to[i]]); 43 ans[bl[g]]=min(ans[bl[g]],d+w[i]+dis[to[i]]); 44 ans[bl[to[i]]]=min(ans[bl[to[i]]],d+w[i]+dis[to[i]]); 45 } 46 if(dis[to[i]]>d+w[i]){ 47 dis[to[i]]=d+w[i];bl[to[i]]=bl[g]; 48 q.push(mmp(-dis[to[i]],to[i])); 49 } 50 } 51 } 52 } 53 inline void work2() 54 { 55 56 57 } 58 int main() 59 { 60 // freopen("distance.in","r",stdin); 61 // freopen("my.out","w",stdout); 62 scanf("%d%d%d",&n,&m,&p); 63 for(int i=1;i<=p;++i)scanf("%d",&a[i]),ans[a[i]]=inf,pd[a[i]]=1; 64 if(p==n){work1();return 0;} 65 for(int i=1,x,y,z;i<=m;++i) 66 { 67 scanf("%d%d%d",&x,&y,&z); 68 addedge(x,y,z);addedge(y,x,z); 69 } 70 getans(); 71 for(int i=1;i<=p;++i) 72 printf("%lld ",ans[a[i]]); 73 }
Analog 92,
T1: a half, a large test sample answer discontinuity found, expression of what is found in three-dimensional partial order, make up the A cdq off.
By being interpreted as a prefix and sort of the discretization b, the array was inserted into the tree array subscripts, complexity nlogn
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #define N 500050 6 #define LL long long 7 using namespace std; 8 int n,ans=1; 9 LL sb[N],lsh[N<<1]; 10 int a[N],b[N],c[N],ls; 11 inline int read(){ 12 int s=0,b=0;char c=getchar(); 13 while(c>'9'||c<'0'){if(c=='-')b=1;c=getchar();} 14 while(c>='0'&&c<='9')s=s*10+c-'0',c=getchar(); 15 if(b)return -s; 16 return s; 17 } 18 struct node{LL a;int b,id;}q[N],qq[N]; 19 inline void init() 20 { 21 sort(lsh+1,lsh+ls+1); 22 ls=unique(lsh+1,lsh+ls+1)-lsh-1; 23 for(int i=0;i<=n;++i) 24 q[i].b=lower_bound(lsh+1,lsh+ls+1,sb[i])-lsh; 25 } 26 inline void add(int x,const int v){ 27 while(x<=ls){ 28 if(c[x]>v)c[x]=v; 29 x+=x&-x; 30 } 31 } 32 inline void del(int x){while(x<=ls){c[x]=n+10;x+=x&-x;}} 33 inline int ask(int x) 34 { 35 int ret=n+10; 36 while(x) 37 { 38 if(c[x]<ret)ret=c[x]; 39 x-=x&-x; 40 } 41 return ret; 42 } 43 inline void CDQ(int l,int r) 44 { 45 if(l==r)return; 46 const int m=l+r>>1; 47 CDQ(l,m);CDQ(m+1,r); 48 register int i=l,j=m+1,t,o=l; 49 while(j<=r) 50 { 51 while(i<=m&&q[j].a>=q[i].a){ 52 add(q[i].b,q[i].id); 53 qq[o++]=q[i++]; 54 } 55 t=q[j].id-ask(q[j].b); 56 if(t>ans)ans=t; 57 qq[o++]=q[j++]; 58 } 59 for(int k=l;k<i;++k)del(q[k].b); 60 while(i<=m)qq[o++]=q[i++]; 61 for(int i=l;i<=r;++i)q[i]=qq[i]; 62 } 63 int main() 64 { 65 scanf("%d",&n); 66 lsh[++ls]=0; 67 for(int i=1,x;i<=n;++i)q[i].a=q[i-1].a+read(),q[i].id=i; 68 for(int i=1,x;i<=n;++i)lsh[++ls]=sb[i]=sb[i-1]+read(),c[i]=n+10; 69 c[0]=n+10; 70 init(); 71 for(int i=n;i<=ls;++i)c[i]=n+10; 72 CDQ(0,n); 73 printf("%d\n",ans); 74 }
T2: violence practices interval dp, dp [i] [j ] represents the answer to the interval [i, j] is, for the length of the enumeration, for the enumeration interval left point, for the root node enumeration complexity of O (n- . 3 )
Optimize the use of decision-making monotonic. It found that the insertion point in an existing section of the right, the root of the original tree is not left. Similarly in the right insertion point is not the right decision.
dp decision points can be recorded simultaneously
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #define N 5050 6 #define LL long long 7 using namespace std; 8 int n,rt=1,po[N][N]; 9 const LL inf=1ll<<50; 10 LL sum[N]; 11 LL ans,dp[N][N]; 12 int main() 13 { 14 scanf("%d",&n); 15 for(int i=1,x;i<=n;++i)scanf("%d",&x),sum[i]=sum[i-1]+x; 16 for(int i=1;i<=n;++i)dp[i][i]=sum[i]-sum[i-1],po[i][i]=i; 17 for(int len=2;len<=n;++len){ 18 for(int i=1,j;i+len-1<=n;++i){ 19 j=i+len-1;dp[i][j]=inf; 20 for(int k=po[i][j-1];k<=po[i+1][j];++k) 21 if(dp[i][k-1]+dp[k+1][j]<=dp[i][j]){dp[i][j]=dp[i][k-1]+dp[k+1][j],po[i][j]=k;} 22 dp[i][j]+=sum[j]-sum[i-1]; 23 } 24 } 25 printf("%lld\n",dp[1][n]); 26 }
T3: Gaussian elimination, the desired title general backstepping. answer:
We first consider a single k. Fi represents the set starting time to the expected number of steps from k i, then
fk = 0, fi = sigma (fj) / xi +1 (i! = k, j for all i the edges, as the degree of i XI), using Gaussian elimination to solve f1.
Specifically partition elimination is: Equation first carried out with elimination of the first half, then the recursive half;
(restored original matrix) performed after elimination half equation reuse, recursively first half. So that when the interval is reduced to
a single point, and this equation is not used to eliminate other equations, we can modify it directly and calculates all fi.
Similar thinking I can refer to this problem solutions: https: //www.cnblogs.com/loadingkkk/p/11272338.html
1 #include<cstdio> 2 #include<iostream> 3 #define N 320 4 #define LL long long 5 using namespace std; 6 const int mod=998244353; 7 int n,m,inv[500050]; 8 LL a[N][N],c[11][N][N],b[N],dd[11][N],ans[N]; 9 inline int qpow(int d,int z) 10 { 11 int ret=1; 12 for(;z;z>>=1,d=1ll*d*d%mod) 13 if(z&1)ret=1ll*ret*d%mod; 14 return ret; 15 } 16 inline void init(int n){for(int i=1;i<=n;++i)inv[i]=qpow(i,mod-2);} 17 int he[N],ne[500050],to[500050],d[N],tot; 18 inline void addedge(int x,int y) 19 { 20 to[++tot]=y;++d[x]; 21 ne[tot]=he[x];he[x]=tot; 22 } 23 inline void Gauss(int l,int r,int x,int y) 24 { 25 for(int i=l;i<=r;++i) 26 { 27 const LL iv=qpow(a[i][i],mod-2); 28 for(int j=1;j<=n;++j) 29 { 30 if(j==i||!a[j][i])continue; 31 const LL pl=iv*a[j][i]%mod; 32 for(int o=x;o<=y;++o){ 33 a[j][o]-=a[i][o]*pl%mod; 34 if(a[j][o]<0)a[j][o]+=mod; 35 } 36 b[j]-=b[i]*pl%mod; 37 if(b[j]<0)b[j]+=mod; 38 } 39 } 40 } 41 inline void solve(int dep,int l,int r) 42 { 43 if(l==r){ans[l]=b[1]*qpow(a[1][1],mod-2)%mod;return;} 44 for(int i=1;i<=n;dd[dep][i]=b[i],++i) 45 for(int j=1;j<=n;++j) 46 c[dep][i][j]=a[i][j]; 47 const int m=l+r>>1; 48 Gauss(l,m,l,r);solve(dep+1,m+1,r); 49 for(int i=1;i<=n;b[i]=dd[dep][i],++i) 50 for(int j=1;j<=n;++j) 51 a[i][j]=c[dep][i][j]; 52 Gauss(m+1,r,l,r);solve(dep+1,l,m); 53 } 54 inline void work() 55 { 56 for(register int i=1;i<=n;++i){ 57 a[i][i]=b[i]=mod-1; 58 for(register int j=he[i];j;j=ne[j]) 59 (a[i][to[j]]+=inv[d[i]])%=mod; 60 } 61 solve(1,1,n); 62 } 63 int main() 64 { 65 // freopen("walk.in","r",stdin); 66 scanf("%d%d",&n,&m);init(m); 67 for(int i=1,x,y;i<=m;++i){ 68 scanf("%d%d",&x,&y); 69 addedge(x,y); 70 } 71 work(); 72 for(int i=2;i<=n;++i)printf("%lld\n",ans[i]); 73 }
I hope next time be able to test well