Given the extent of the brutality T3, I decided to first take a decadent piece solution to a problem.
T1 urban (city)
Inclusion and exclusion method baffle +
Construction teams of m into n groups, each group must have a, is not considered an upper limit, a total of C (m-1, n-1) kinds of programs.
I-th groups of more than k, off-repellent capacity C (n, i) * C (mi * k-1, n-1) × k i corresponds to a construction teams out, the remaining mi * construction of k into n groups and teams to ensure that each group has at least one and no consideration bound, then this i-k into n packets where there are at least two groups is greater than i k.
#include<iostream> #include<cstring> #include<cstdio> #define mod 998244353 #define ll long long using namespace std; ll n,m,k,fac[10001000],inv[10001000],facinv[10001000],ans; ll C(ll x,ll y) { if(y>x) return 0; return fac[x]*facinv[y]%mod*facinv[x-y]%mod; } int main() { scanf("%lld%lld%lld",&n,&m,&k); if(n>m){ puts("0"); return 0; } fac[0]=1;facinv[0]=1;inv[1]=1; for(int i=1;i<=max(n,m);i++){ if(i!=1) inv[i]=(mod-mod/i)*inv[mod%i]%mod; fac[i]=fac[i-1]*i%mod; facinv[i]=facinv[i-1]*inv[i]%mod; } for(int i=1;i<=n;i++){ if(i&1) ans=(ans+C(n,i)*C(m-i*k-1,n-1))%mod; else ans=(ans-C(n,i)*C(m-i*k-1,n-1)+mod)%mod; } printf("%lld\n",(C(m-1,n-1)-ans+mod)%mod); return 0; }
T2 bombing (bomb)
Exam wrong a lot of questions, but I did not wrong to do it (is a waste)
As long as there is movement path can reach, then we point tarjan shrink, shrink after completion FIG build a new point, to find the longest chain, the depth of each point should be the parent node point + depth size (a few points lump together ), topu request longest chain, or dfs (dfs but seems to be easy T).
#include<iostream> #include<cstdio> #include<queue> using namespace std; struct node { int to,nxt; }h[4001000],hh[4001000]; int n,m,nxt[4001000],tot,tet,cnt,dfn[1001000],low[1001000],s[1001000]; int top,num,whos[1001000],sz[1001000],du[1001000],dep[1001000],ans,nx[4001000]; bool is[1001000]; int max ( int x, int y) { return x> y? x: y; } Void add ( int x, int y) { h [ ++ tot] .to = y; h [tot] .nxt = nxt [x]; nxt [x] = tot; } Void product ( int x, int y) { HH [ ++ tet] .to = y; HH [tet] .nxt = NX [x]; NX [x] = tet; } Void tarjan ( int x) { dfn[x]=low[x]=++cnt; s[++top]=x;is[x]=1; for(int i=nxt[x];i;i=h[i].nxt){ int y=h[i].to; if(!dfn[y]){ tarjan(y); low[x]=min(low[x],low[y]); } else if(is[y]) low[x]=min(low[x],dfn[y]); } if(dfn[x]==low[x]){ num++; while(1){ int tmp=s[top--]; is[tmp]=0; whos[tmp]=num; sz[num]++; if(x==tmp) break; } } } void topu() { queue<int>q; for(int i=1;i<=num;i++) if(du[i]==0) q.push(i),dep[i]=sz[i],ans=max(ans,dep[i]); while(q.size()){ int x=q.front();q.pop(); for(int i=nx[x];i;i=hh[i].nxt){ int y=hh[i].to; du[y]--; dep[y]=max(dep[y],dep[x]+sz[y]); ans=max(ans,dep[y]); if(!du[y]) q.push(y); } } } int main() { scanf("%d%d",&n,&m); int u,v; for(int i=1;i<=m;i++){ scanf("%d%d",&u,&v); add(u,v); } for(int i=1;i<=n;i++) if(!dfn[i]) tarjan(i); for(int i=1;i<=n;i++){ for(int j=nxt[i];j;j=h[j].nxt){ int y=h[j].to; if(whos[i]!=whos[y]){ ad(whos[i],whos[y]); du[whos[y]]++; } } } topu(); printf("%d\n",ans); }