# 考前模版整理

ll pow(ll n, ll t)
{
ll ret = 1;
for(; t; t >>= 1, n = (n * n) % mo)
if(t & 1) ret = (ret * n) % mo;
return ret;
}

ll inv(ll n, ll p)
{
ll ret = 1;
for(ll i = p; i; i >>= 1, n = (n * n) % mo)
if(i & 1) ret = (ret * n) % mo;
return ret;
}
int main()
{
inv(b,mo - 2)
//a/b，求b对mo的逆元
}

ll mul(ll x, ll y)
{
ll ret = 0;
for(; y; y >>= 1, x = (x + x) % P)
if(y & 1) ret = (ret + x) % P;
return ret;
}

int heap[10010], heap_size;
void put(int d)
{
int now, next;
heap[++heap_size] = d;
now = heap_size;
while(now > 1)
{
next = now >> 1;
if(heap[now] >= heap[next]) return;
swap(heap[now], heap[next]);
now = next;
}
}
int get()
{
int res = heap[1], now, next;
heap[1] = heap[heap_size--];
now = 1;
while(now * 2 <= heap_size)
{
next = now << 1;
if(next < heap_size && heap[next] > heap[next|1]) next|=1;
if(heap[next] >= heap[now]) break;
swap(heap[next], heap[now]);
now = next;
}
return res;
}

struct Que
{
int head, tail, ck[10000 + 3];
void fresh() {head = 1; tail = 0;}
void pop_top(RG int p)
void push(RG int i, RG ll k, RG int p)
{
while(head <= tail && f[i - 1][ck[tail]] - G[i] * X[ck[tail]] > k) --tail;
ck[++tail] = p;
}
}Q;

	for(RG int i = 1; i <= n; ++i)
{
if(a[i] > d[len]) d[++len] = a[i], f[i] = len;
else
{
RG int L = 1, R = len, mid;
while(L <= R)
{
mid = L + R >> 1;
if(d[mid] >= a[i]) R = mid - 1;
else L = mid + 1;
}
d[L] = a[i];
f[i] = L;
}
}
len = 0, d[0] = oo;
for(RG int i = n; i; --i)
{
if(a[i] < d[len]) d[++len] = a[i], g[i] = len;
else
{
RG int L = 1, R = len, mid;
while(L <= R)
{
mid = L + R >> 1;
if(d[mid] <= a[i]) R = mid - 1;
else L = mid + 1;
}
d[L] = a[i];
g[i] = L;
}
}


#include <queue>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define oo 0x7fffffff - 1
#define max(a, b) ((a) > (b) ? (a) : (b))
using namespace std;
{
int c = getchar(), x = 0;
while(c < '0' || c > '9') c = getchar();
while(c >= '0' &&  c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
return x;
}
int ans, n, m, A, B, C, D, cnt, ncnt, fcnt, h[50003], nh[50003], fh[50003], d1[50003], d2[50003], hash[200003], du[50003], Q[50003], f[50003];
struct pt {int v, w, ne;} a[200003], na[200003], fa[200003];
struct ed {int u, v, w;} e[200003];
struct abcd {
int fir, sec;
bool operator < (const abcd oth) const {return sec > oth.sec;} };
priority_queue<abcd> q;
bool mark[50003];
inline void link(int u, int v, int w) {a[++cnt].v = v, a[cnt].w = w, a[cnt].ne = h[u], h[u] = cnt;}
inline void nlink(int u, int v, int w) {na[++ncnt].v = v, na[ncnt].w = w, na[ncnt].ne = nh[u], nh[u] = ncnt;}
inline void flink(int u, int v, int w) {fa[++fcnt].v = v, fa[fcnt].w = w, fa[fcnt].ne = fh[u], fh[u] = fcnt;}
void dijkstra(int x, int d[], pt a[], int h[])
{
for(int i = 1; i <= n; ++i) d[i] = oo, mark[i] = 0;
d[x] = 0; q.push((abcd){x, 0});
while(!q.empty())
{
int now = q.top().fir; q.pop();
if(mark[now]) continue;
mark[now] = 1;
for(int j = h[now]; j; j = a[j].ne)
if(!mark[a[j].v] && d[a[j].v] > d[now] + a[j].w)
{
d[a[j].v] = d[now] + a[j].w;
q.push((abcd){a[j].v, d[a[j].v]});
}
}
}
bool preced()
{
dijkstra(A, d1, a, h);
dijkstra(B, d2, na, nh);
if(d1[B] == oo || d2[A] == oo) return 0;
for(int j = 1; j <= m; ++j) if(d1[e[j].u] + e[j].w + d2[e[j].v] == d1[B]) ++hash[j];
dijkstra(C, d1, a, h);
dijkstra(D, d2, na, nh);
if(d1[D] == oo || d2[C] == oo) return 0;
for(int j = 1; j <= m; ++j)
{
if(d1[e[j].u] + e[j].w + d2[e[j].v] == d1[D]) ++hash[j];
if(hash[j] == 2) {flink(e[j].u, e[j].v, e[j].w); ++du[e[j].v];}
}
return 1;
}
void sortdp()
{
for(int i = 1; i <= n; ++i) {f[i] = 1; if(!du[i]) Q[++Q[0]] = i;}
for(int i = 1; i <= Q[0]; ++i)
{
int now = Q[i];
for(int j = fh[now]; j; j = fa[j].ne)
{
f[fa[j].v] = max(f[fa[j].v], f[now] + 1);
ans = max(ans, f[fa[j].v]);
--du[fa[j].v];
if(!du[fa[j].v]) Q[++Q[0]] = fa[j].v;
}
}
printf("%d\n", ans);
}
int main()
{
freopen("game.in", "r", stdin), freopen("game.out", "w", stdout);
for(int i = 1; i <= m; ++i)
{
}
if(preced()) sortdp();
else puts("-1");
fclose(stdin), fclose(stdout);
return 0;
} 

#include<stdio.h>
#pragma GCC optimize("O2")
using namespace std;
#define inline __inline__ __attribute__((always_inline))
namespace io{
const int MAXBUF=1<<9;
char B[MAXBUF],*S=B,*T=B;
#define gt() (S == T && (T = (S = B) + fread(B, 1, MAXBUF, stdin), S == T) ? 0 : *S++)
Type aa=0;bool bb=0;char c,*S=io::S,*T=io::T;
for(c=gt();(c<'0'||c>'9')&&c!='-';c=gt())
;
for(c=='-'?bb=1:aa=c-'0',c=gt();c>='0'&&c<='9';c=gt())
aa=(aa<<1)+(aa<<3)+c-'0';
io::S=S,io::T=T;return bb?-aa:aa;
}
char BUF[MAXBUF],*iter=BUF;
template<class T> void P(T x,char c='\n'){
static int stk[110];int O=0;char *iter=io::iter;
if(!x)*iter++='0';
else{
if(x<0)x=-x,*iter++='-';
for(;x;x/=10)stk[++O]=x%10;
for(;O;*iter++='0'+stk[O--])
;
}
*iter++=c,io::iter=iter;
}
inline void output(){
fwrite(BUF,1,iter-BUF,stdout),iter=BUF;
}
}
namespace pb_ds{
const int N=500001;
struct Pointer{int to;Pointer *nxt;}*fst[N];
static Pointer mem[N<<1],*tot=mem;
*++tot=(Pointer){v,fst[u]},fst[u]=tot;
*++tot=(Pointer){u,fst[v]},fst[v]=tot;
}
bool vis[N];
int fa[N],dep[N],sz[N],top[N],hs[N];
void dfs_init(int x){
vis[x]=sz[x]=1;
for(Pointer *iter=fst[x];iter;iter=iter->nxt)
if(!vis[iter->to]){
fa[iter->to]=x;
dep[iter->to]=dep[x]+1;
dfs_init(iter->to);
sz[x]+=sz[iter->to];
if(sz[hs[x]]<sz[iter->to])
hs[x]=iter->to;
}
}
void dfs_make(int x){
vis[x]=0;
top[x]=x^hs[fa[x]]?x:top[fa[x]];
if(hs[x]){
dfs_make(hs[x]);
for(Pointer *iter=fst[x];iter;iter=iter->nxt)
if(vis[iter->to])
dfs_make(iter->to);
}
}
inline int lca(int x,int y){
while(top[x]^top[y])
dep[top[x]]>dep[top[y]]
?x=fa[top[x]]
:y=fa[top[y]];
return dep[x]<dep[y]?x:y;
}
void main(){
int n=F(),que=F(),root=F();
for(int i=1,x,y;i<n;i++)
dfs_init(root);
dfs_make(root);
for(int x,y;que;que--)
x=F(),y=F(),io::P(lca(x,y));
}
}
int main(){
pb_ds::main();
io::output();
}


posted @ 2017-10-19 17:56  keshuqi  阅读(215)  评论(0编辑  收藏  举报