【组队训练】2016 ACM/ICPC Asia Regional Dalian Online

因为不是一队……毫无晋级的压力……反正有压力也进不去呵呵呵……

开场zr看1006我看1010。。

1010我一直在wa。。。

zr的1006倒是比较轻松的过了。。。然后我让他帮我看10。。。。

跟他讲了半天我代码的逻辑。。。然后我自己看出来的。。。。比赛的代码。。。。写的十分混乱。。。。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
typedef long long ll;
using namespace std;
const int N = ;
vector<int> G[N];

int a[N], b[N];
int low[N];
int par[N];
int c[N];
int cntv;
ll ans;
int n;

int lowbit(int x)
{
    return x & (-x);
}

int sum(int n)                //前n个数的和
{
    int ans = ;
    while (n > ) {
        ans +=  c[n];
        n -= lowbit(n);
    }
    return ans;
}

void add(int pos, int num, int n)    //在pos处加num
{
    while (pos <= n) {
        c[pos] += num;
        pos += lowbit(pos);
    }
}

void dfs(int u) {
    //printf("==%d %d %d\n", u, a[u], low[ a[u] ]);
    //printf("n = %d\n", n);
    int lowu = low[ a[u] ];
    int tmp = sum(lowu);
    for (unsigned i = ; i < G[u].size(); ++i) {
        int v = G[u][i];
        dfs(v);
    }
    int tmp2 = sum(lowu);
    add(a[u], , n);
    cntv++;
    //printf(">>%d %d %d\n", u, tmp, tmp2);
    ans += tmp2 - tmp;
}

void init() {
    memset(c, , sizeof c);
    memset(low, , sizeof low);
    memset(par, , sizeof par);
}

int main()
{
    //freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        ll k;
        init();
        scanf("%d%lld", &n, &k);
        //printf("%d %lld\n", n, k);
        for (int i = ; i <= n; ++i) {
            G[i].clear();
            scanf("%d", &a[i]);
            b[i] = a[i];
        }
        sort(b+, b++n);
        int cnt = unique(b+, b++n) - (b+);
        //printf("cnt=%d\n",cnt);
        //for (int i = 1; i <= cnt; ++i) printf("|%d ", b[i]); printf("\n");
        int idx = cnt;
        for (int i = ; i <= cnt; ++i) {
            while ((ll)b[i]*b[idx--] > k) ;
            idx++; low[i] = idx;
        }//low[i]记录的是i*low[i]<=k的最大值
        //for (int i = 1; i <= n; ++i) printf("%d ", low[i]); printf("\n");
        for (int i = ; i <= n; ++i) {
            a[i] = lower_bound(b+, b++cnt, a[i]) - b;
        }
        //for (int i = 1; i <= n; ++i) { printf("%d %d\n", a[i], low[ a[i] ]); }

        int u, v;
        for (int i = ; i < n; ++i) {//printf("i=%d\n", i);
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
            par[v] = u;
        }
        int root = ;
        while (par[root]) root = par[root];
        //printf("root=%d\n", root);
        cntv = ;
        ans = ;
        dfs(root);
        printf("%lld\n", ans);
    }
    return ;
}

1009总体思路也是我想出来的，但是代码出了一些十分隐蔽的逻辑错误。。。zr比较强大。。。竟然看出来并改对了。。。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
typedef long long ll;
using namespace std;
const int N = ;
vector<int> G[N];
int ans[N];

void up(int &x, int y) {
    if (x == - || x > y+) x = y+;
}

int main()
{//freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        int n, m;
        scanf("%d%d", &n, &m);
        for (int i = ; i <= n; ++i) G[i].clear();
        int u, v;
        for (int i = ; i < m; ++i) {
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        int s;
        scanf("%d", &s);
        int cnt = ;
        for (int i = ; i <= n; ++i) ans[i] = ; ans[s] = -;
        queue<int> q[];
        int now = ;
        for (unsigned i = ; i < G[s].size(); ++i) {
            int v = G[s][i];
            ans[v] = -;
            q[now].push(v);
            cnt++;
        }
        for (int i = ; i <= *m; ++i) {
            if (q[now].empty()) break;
            int num = ;
            while (q[now].size()) {
                int u = q[now].front(); q[now].pop();
                //printf("u=%d \n", u);
                int tmp = ;
                int sz = G[u].size();
                for (int j = ; j < sz; ++j) {
                    int v = G[u][j];
                    if (ans[v] == -) tmp++;
                }
                //printf("%d: %d %d %d\n", u, sz, tmp, n);
                if (n - sz > cnt - tmp) {
                    //printf(">>%d %d\n", u, ans[u]);
                    ans[u] = i+;
                    num++;
                } else {
                    q[now^].push(u);
                }

            }
            cnt -= num;
            now ^= ;
        }

        bool ok = true;
        for (int i = ; i <= n; ++i) {
            if (s == i) continue;
            if (ok) ok = false; else printf(" ");
            printf("%d", ans[i]);
        }
        printf("\n");
    }
    return ;
}

1007坑题。竟然是推公式。。。两个人讨论了下，wa了几次推出来了。。。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <iostream>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
typedef long long ll;
#define CLR(x, v) sizeof (x, v, sizeof(x))
using namespace std;
ll n,m;
int main()
{
   //freopen("in.txt","r",stdin);
   while(cin >> m >> n){
        if (m == ) puts("T");
        else if ((m/)*(m-m/) > n) puts("F");
        else puts("T");
   }
   return ;
}

02补题吧。。太弱了。。。。