BestCoder Round #72 (div.2)

后面的题目补不懂了

暴力 1001 Clarke and chemistry

这题也把我搞死了。。枚举系数判断就行了

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>

int cnt[3][30];

bool error(void)    {
    for (int i=0; i<2; ++i) {
        for (int j=0; j<26; ++j)    {
            if (cnt[i][j] != -1 && cnt[2][j] == -1) {
                return true;
            }
        }
    }
    return false;
}

int main(void)  {
    int T;  scanf ("%d", &T);
    while (T--) {
        int A, B, C;    scanf ("%d%d%d", &A, &B, &C);
        memset (cnt, -1, sizeof (cnt));
        char c[2];    int t;
        for (int i=0; i<A; ++i) {
            scanf ("%s %d", &c, &t);
            cnt[0][c[0]-'A'] = t;
        }
        for (int i=0; i<B; ++i) {
            scanf ("%s %d", &c, &t);
            cnt[1][c[0]-'A'] = t;
        }
        for (int i=0; i<C; ++i) {
            scanf ("%s %d", &c, &t);
            cnt[2][c[0]-'A'] = t;
        }
        bool flag = true;
        int ans1 = 1000000, ans2 = 1000000;
        for (int i=1; i<=2000&&flag; ++i) {
            for (int j=1; j<=2000&&flag; ++j) {
                bool ok = true;
                for (int k=0; k<26; ++k) {
                    if (cnt[2][k] == -1)    continue;
                    if (cnt[0][k] == -1 && cnt[1][k] == -1) {
                        flag = false;   break;
                    }
                    int x = 0;
                    if (cnt[0][k] != -1)    x = cnt[0][k] * i;
                    if (cnt[1][k] != -1)    x += cnt[1][k] * j;
                    if (x != cnt[2][k]) {
                        ok = false; break;
                    }
                }
                if (ok) {
                    if (i < ans1 || (i == ans1 && j < ans2))   ans1 = i, ans2 = j;
                }
            }
        }
        if (error ())  flag = false;
        if (flag && ans1 < 1000000)   printf ("%d %d\n", ans1, ans2);
        else    puts ("NO");
    }

    return 0;
}

数学 1002 Clarke and points

题意： 求|XA - XB| + |YA - YB| 最大

分析：去掉绝对值，就知道只要得到最大最小的(XA + XB) 和 (XA - XB)

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <ctime>
#include <cstdlib>
#include <cmath>
#include <iostream>
using namespace std;

long long seed;
inline long long rand(long long l, long long r) {
    static long long mo=1e9+7, g=78125;
    return l+((seed*=g)%=mo)%(r-l+1);
}

const int N = 1e6 + 5;
pair<long long, long long> p[N];
long long mx[2], mn[2];

int main(void)  {
    int T;  cin >> T;
    while (T--) {
        int n;
        cin >> n >> seed;
        mx[0] = mx[1] = -(1ll << 60);
        mn[0] = mn[1] = (1ll << 60);
        for (int i = 0; i < n; i++) {
            p[i].first = rand (-1000000000, 1000000000);
            p[i].second = rand (-1000000000, 1000000000);
            mx[0] = max (mx[0], p[i].first + p[i].second);
            mx[1] = max (mx[1], p[i].first - p[i].second);
            mn[0] = min (mn[0], p[i].first + p[i].second);
            mn[1] = min (mn[1], p[i].first - p[i].second);
        }
        cout << max (abs (mx[0] - mn[0]), abs (mx[1] - mn[1])) << '\n';
    }
    return 0;
}

贪心 + BFS Clarke and MST

题意：求位运算and的最大生成树

分析：枚举数字每一位是否有可能为1，即(now & w == now)，用BFS遍历所有生成树

#include <cstdio>
#include <cstring>
#include <queue>

const int N = 3e5 + 5;
struct Edge {
    int v, w;
};
std::vector<Edge> G[N];

int n, m;
bool vis[N];

bool BFS(int now)   {
    std::queue<int> que;
    memset (vis, false, sizeof (vis));
    que.push (1);   vis[1] = true;
    while (!que.empty ())   {
        int u = que.front ();   que.pop ();
        for (int i=0; i<G[u].size (); ++i)  {
            int v = G[u][i].v;
            int w = G[u][i].w;
            if (!vis[v] && (w & now) == now)    {
                vis[v] = true;
                que.push (v);
            }
        }
    }
    for (int i=1; i<=n; ++i)    if (!vis[i])    return false;
    return true;
}

int main(void)  {
    int T;  scanf ("%d", &T);
    while (T--) {
        scanf ("%d%d", &n, &m);
        for (int i=0; i<=n; ++i)    G[i].clear ();
        for (int u, v, w, i=0; i<m; ++i)   {
            scanf ("%d%d%d", &u, &v, &w);
            G[u].push_back ((Edge) {v, w});
            G[v].push_back ((Edge) {v, w});
        }
        int ans = 0;
        for (int i=30; i>=0; --i)   {
            int now = ((1 << i) | ans);
            if (BFS (now))  ans = now;
        }
        printf ("%d\n", ans);
    }

    return 0;
}