美团CodeM复赛 02,03

02 城市网络
比赛时候写的是单调栈，真的是让人见笑了，基本思路就是dfs时候动态处理单调栈(带回溯)，然后离线处理答案。题解是用了倍增的，效率高很多

#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <algorithm>#include <iostream>#include <map>#include <set>#include <queue>#include <cmath>using namespace std;typedef long long ll;#define MP(x, y) make_pair(x, y)#define lson l,m, rt<<1#define rson m+1, r, rt<<1|1const int N = 1e5+5;const int INF = 0x3f3f3f3f;int A[N];struct Node{    int to, nx;}E[N * 2];int Deep[N];int head[N], tot;void add(int fr, int to) {    E[tot].to = to; E[tot].nx = head[fr]; head[fr] = tot ++;}struct Pode{    int po, val, an;    Pode(int a=0, int b=0, int c=0):po(a), val(b), an(c){}};vector<Pode> ask[N];int ans[N];int SA[N]; int SB[N]; int cnt;int search(int x, int ty) {    int l = 1; int r = cnt;    while(l <= r) {        int mid = (l + r) >> 1;        if(ty) {          if(SA[mid] >= x) r = mid - 1;            else l = mid + 1;        }else {            if(-SB[mid] >= -x) r = mid-1;            else l = mid + 1;        }    }    return l;}void dfs(int x, int pre, int dep) {    Deep[x] = dep;    vector<pair<int, int> > tmp;    if(cnt == 0) SA[++cnt] = dep, SB[cnt] = A[x];    else {        int pos = 0;        for(int i = cnt; i >= 1; --i) {            if(SB[i] > A[x]) {                pos = i;                SA[i + 1] = dep; SB[i + 1] = A[x]; cnt = i+1;                break;            }else {                tmp.push_back(MP(SA[i], SB[i]));            }        }        if(pos == 0) SA[1] = dep, SB[1] = A[x], cnt = 1;    }//  printf("%d: ", x); for(int i = 1; i <= cnt; ++i) printf("%d:%d ", SA[i], SB[i]); printf("\n");    for(int i = 0; i < ask[x].size(); ++i) {        int po = ask[x][i].po; int val = ask[x][i].val; int an = ask[x][i].an;        int pos1 = search(Deep[po], 1); int pos2 = search(val, 0);        ans[an] = max(pos2 - pos1, 0);    }    for(int i = head[x]; ~i; i = E[i].nx) {        int to = E[i].to; if(to == pre) continue;        dfs(to, x, dep + 1);    }    cnt --;    for(int i = tmp.size() - 1; i >= 0; --i) {        int t1 = tmp[i].first; int t2 = tmp[i].second;        SA[++cnt] = t1; SB[cnt] = t2;     }    tmp.clear();}int main() {    int n, q;    while(~scanf("%d %d", &n, &q)) {        cnt = 0;        memset(head, -1, sizeof(head)); tot = 0;        for(int i = 1; i <= n; ++i) ask[i].clear();        for(int i = 1; i <= n; ++i) scanf("%d", &A[i]);        for(int i = 1; i < n; ++i) {            int a, b; scanf("%d %d", &a, &b);            add(a, b); add(b, a);        }        for(int i = 0; i < q; ++i) {            int a, b, c; scanf("%d %d %d", &a, &b, &c);            ask[a].push_back(Pode(b, c, i));        }        dfs(1, 1, 1);    //  for(int i = 1; i <= n; ++i) printf("%d ", Deep[i]); printf("\n");        for(int i = 0; i < q; ++i) printf("%d\n", ans[i]);    }    return 0;}

03 神秘代码
比赛最后10分钟我才想到这就是一个环套树，然后就没有然后了= =，如果能A的话还是很有希望进前50的，我之前认为这题看起来像数论，又那么像拓展欧几里得，不会不会，还不如多试试05。确实弱233

#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <algorithm>#include <iostream>#include <map>#include <set>#include <queue>#include <cmath>using namespace std;typedef long long ll;#define MP(x, y) make_pair(x, y)#define lson l,m, rt<<1#define rson m+1, r, rt<<1|1const int N = 1e5+5;const int INF = 0x3f3f3f3f;int MOD;struct Node{    int fr, to, nx, a, b, c;}E[N << 1];int head[N]; int tot;int vis[N]; int Pre[N];int suc;int ans[N];void add(int fr, int to, int a, int b, int c) {    E[tot].to = to; E[tot].a = a; E[tot].b = b; E[tot].c = c; E[tot].fr = fr;    E[tot].nx = head[fr]; head[fr] = tot ++;}long long inv(long long a) {    if(a == 1) return 1;    return inv(MOD%a)*(MOD-MOD/a)% MOD; }void bfs(int x) {    queue<int> Q;    Q.push(x);    vis[x] = 2;    while(!Q.empty()) {        int po = Q.front(); Q.pop();    //  printf("%d: %d\n", po, ans[po]);        for(int i = head[po]; ~i; i = E[i].nx) {            int y = E[i].to; int a = E[i].a; int b = E[i].b; int c = E[i].c;            if(vis[y] != 2) {                vis[y] = 2;                ll tt = inv(b);                ans[y] = (1ll*c*tt %MOD - 1ll*a*ans[po]%MOD * tt %MOD + MOD) %MOD;                Q.push(y);            }        }    }}void Find(int x, int tar, int B, int C) {    int Next = E[Pre[x]].fr; int a = E[Pre[x]].b; int b = E[Pre[x]].a; int c = E[Pre[x]].c;    ll tmp = inv(a);    //  printf("a:%d b:%d c:%d Next:%d now:%d %lld %d %d\n", a,b,c, Next, x, tmp, B, C);    b = 1ll* (-b+MOD) * tmp %MOD;    c = 1ll* c * tmp %MOD;    C = (1ll*C + 1ll*B*c) % MOD;    B = 1ll* B*b %MOD;    if(Next == tar) {        ans[tar] = 1ll* C*inv( (1-B+MOD)%MOD ) % MOD;        //  printf("tar: %d\n", ans[tar]);        bfs(tar);        return;     }    Find(Next, tar, B, C);}void dfs(int x, int pre) {    if(suc) return;    for(int i = head[x]; ~i; i = E[i].nx) {        int to = E[i].to; if(to == pre) continue;        if(suc) return;        if(!vis[to]) { vis[to] = 1; Pre[to] = i; dfs(to, x); }        else {            Pre[to] = i;            Find(to, to, 1, 0);            suc = 1;             break;        }    }}int main() {    int n;    while(~scanf("%d %d", &n, &MOD)) {        memset(head, -1, sizeof(head));        tot = 0;        for(int i = 0; i < n; ++i) {            int u, v, a, b, c;            scanf("%d %d %d %d %d", &u, &v, &a, &b, &c);            add(u, v, a, b, c);            add(v, u, b, a, c);        }        for(int i = 1; i <= n; ++i) {            if(!vis[i]) {                suc = 0;                vis[i] = 1; dfs(i, i);            }        }        for(int i = 1; i <= n; ++i) printf("%d\n", ans[i]);    }       return 0;}