2016 ACM/ICPC亚洲区青岛站

 

[A. Relic Discovery]

签到

#include 

int n;

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        int ans = 0;
        for (int i = 0; i < n; i++) {
            int a, b;
            scanf("%d%d", &a, &b);
            ans += a * b;
        }
        printf("%d\n", ans);
    }
    return 0;
}

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[B. Pocket Cube]

判断所有情况，写丑了。

#include 

const int N = 30;
int a[N], b[N];
const int n = 24;

bool check() {
    for (int i = 1; i <= 6; i++) {
        int j = (i - 1) * 4 + 1;
        int num = b[j];
        for (int cnt = 0; cnt < 4; j++, cnt++)
            if (b[j] != num) return false; 
    }
    return true;
}

int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        for (int i = 1; i <= 24; i++)
            scanf("%d", a + i);
        memcpy(b, a, sizeof(a));
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(a));
        b[5] = a[1]; b[7] = a[3];
        b[9] = a[5]; b[11] = a[7];
        b[13] = a[9]; b[15] = a[11];
        b[3] = a[15]; b[1] = a[13];
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(b));
        b[1] = a[5]; b[3] = a[7];
        b[5] = a[9]; b[7] = a[11];
        b[9] = a[13]; b[11] = a[15];
        b[15] = a[3]; b[13] = a[1];
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(b));
        b[6] = a[2]; b[8] = a[4];
        b[10] = a[6]; b[12] = a[8];
        b[14] = a[10]; b[16] = a[12];
        b[4] = a[16]; b[2] = a[14];
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(b));
        b[4] = a[8]; b[2] = a[6];
        b[6] = a[10]; b[8] = a[12];
        b[10] = a[14]; b[12] = a[16];
        b[14] = a[2]; b[16] = a[4];
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(b));
        b[3] = a[19]; b[4] = a[20];
        b[23] = a[3]; b[24] = a[4];
        b[9] = a[24]; b[10] = a[23];
        b[20] = a[9]; b[19] = a[10];
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(b));
        b[3] = a[23]; b[4] = a[24];
        b[23] = a[10]; b[24] = a[9];
        b[10] = a[19]; b[9] = a[20];
        b[19] = a[3]; b[20] = a[4];
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(b));
        b[1] = a[21]; b[2] = a[22];
        b[21] = a[12]; b[22] = a[11];
        b[12] = a[17]; b[11] = a[18];
        b[17] = a[1]; b[18] = a[2];
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(b));
        b[1] = a[17]; b[2] = a[18];
        b[21] = a[1]; b[22] = a[2];
        b[11] = a[22]; b[12] = a[21];
        b[18] = a[11]; b[17] = a[12];
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(b));
        b[5] = a[23]; b[6] = a[21];
        b[23] = a[16]; b[21] = a[15];
        b[15] = a[20]; b[16] = a[18];
        b[20] = a[6]; b[18] = a[5];
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(b));
        b[5] = a[18]; b[6] = a[20];
        b[23] = a[5]; b[21] = a[6];
        b[16] = a[23]; b[15] = a[21];
        b[20] = a[15]; b[18] = a[16];
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(b));
        b[7] = a[17]; b[8] = a[19];
        b[24] = a[7]; b[22] = a[8];
        b[13] = a[22]; b[14] = a[24];
        b[19] = a[13]; b[17] = a[14];
        if (check()) {
            puts("YES");
            continue;
        }
        memcpy(b, a, sizeof(b));
        b[7] = a[24]; b[8] = a[22];
        b[24] = a[14]; b[22] = a[13];
        b[13] = a[19]; b[14] = a[17];
        b[19] = a[8]; b[17] = a[7];
        if (check()) {
            puts("YES");
            continue;
        }
        puts("NO");
    }
    return 0;
}

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[C. Pocky]

当 $n > d$ 时，$f(n) = \dfrac{\int_{d} ^{n} f(x)dx}{n} + 1$
当 $n \leq d$ 时，$f(n) = 0$
解得 $f(n) = ln(\dfrac{n}{d}) + 1$

#include

const double eps = 1e-10;

int dcmp(double x) {
    if (fabs(x) < eps) return 0;
    return x < 0 ? -1 : 1;
}

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        double a, b;
        scanf("%lf%lf", &a, &b);
        if (dcmp(a - b) <= 0) {
            puts("0.000000");
            continue;
        }
        double ans = log(a) - log(b) + 1.0;
        printf("%.6f\n", ans);
    }
    return 0;
}

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[D. Lucky Coins]

每种硬币都是独立的。那么求出第 $i$ 种硬币在第 $j$ 轮之前都被拿掉的概率 $die[i][j] = (1 - p_{i} ^ j)^{cnt_i}$。$alive[i][j] = 1 - die[i][j]$ 表示第 $i$ 种硬币在第 $j$ 轮还存活的概率。
对于第 $i$ 种硬币的答案就是 $\sum_{step} (alive[i][step] - alive[i][step + 1]) * \prod_{j!=i}die[j][step]$。

#include 

const int N = 15;
const int step = 105;

double qp(double a, int n) {
    double ans = 1;
    while (n) {
        if (n & 1) ans *= a;
        a *= a;
        n >>= 1;
    }
    return ans;
}

double alive[N][step], die[N][step], p[N];
int cnt[N], n;

void init(int index) {
    double pp = p[index];
    for (int i = 1; i < step; i++) {
        die[index][i] = qp(1 - pp, cnt[index]);
        alive[index][i] = 1 - die[index][i];
        pp *= p[index];
    }
}

double solve(int index) {
    double ans = 0;
    for (int i = 1; i < step - 1; i++) {
        double temp = alive[index][i] - alive[index][i + 1];
        for (int j = 1; j <= n; j++) {
            if (j == index) continue;
            temp *= die[j][i];
        }
        ans += temp;
    }
    return ans;
}

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%d%lf", cnt + i, p + i);
            init(i);
        }
        if (n == 1) {
            puts("1.000000");
            continue;
        }
        for (int i = 1; i <= n; i++) 
            printf("%.6f%c", solve(i), " \n"[i == n]);
    }
    return 0;
}

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[G. Coding Contest]

求出最小的崩溃概率，即求最大的未崩溃概率。

转化成二分图。人多于食物的为 $X$，人少于食物的为 $Y$。

概率取对数，跑费用流即可。

#include 

const int N = 500 + 7;
const int M = 2e4 + 7;
const double inf = 1e12;
const int INF = 0x3f3f3f3f;

struct E {
    int v, ne, f;
    double c;
} e[M];
int head[N], cnt;
double dis[N];
int path[N], n, m;
bool inq[N];

inline void add(int u, int v, int f, double c) {
    e[cnt].v = v; e[cnt].f = f; e[cnt].c = c; e[cnt].ne = head[u]; head[u] = cnt++;
    e[cnt].v = u; e[cnt].f = 0; e[cnt].c = -c; e[cnt].ne = head[v]; head[v] = cnt++;
}

const double eps = 1e-8;

int dcmp(double x) {
    if (fabs(x) < eps) return 0;
    return x < 0 ? -1 : 1;
}

bool spfa(int s, int t) {
    for (int i = 0; i <= t; i++)
        dis[i] = inf, inq[i] = 0, path[i] = -1;
    dis[s] = 0;
    inq[s] = 1;
    std::queue<int> que;
    que.push(s);
    while (!que.empty()) {
        int u = que.front(); que.pop();
        inq[u] = 0;
        for (int i = head[u]; ~i; i = e[i].ne) {
            int v = e[i].v; double c = e[i].c;
            if (e[i].f && dcmp(dis[v] - dis[u] - c) > 0) {
                dis[v] = dis[u] + c;
                path[v] = i;
                if (!inq[v]) {
                    inq[v] = 1;
                    que.push(v);
                }
            }
        }
    }
    return dis[t] != inf;
}

double mcf(int s, int t) {
    double ans = 0;
    while (spfa(s, t)) {
        int x = INF;
        for (int i = path[t]; ~i; i = path[e[i ^ 1].v]) x = std::min(x, e[i].f);
        ans += dis[t] * x;
        for (int i = path[t]; ~i; i = path[e[i ^ 1].v]) e[i].f -= x, e[i ^ 1].f += x;
    }
    return ans;
}

int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        memset(head, -1, sizeof(head));
        cnt = 0;
        scanf("%d%d", &n, &m);
        int s = 0, t = n + 1;
        for (int i = 1; i <= n; i++) {
            int S, B;
            scanf("%d%d", &S ,&B);
            if (S > B) add(s, i, S - B, 0);
            else if(S0);
        }
        for (int i = 1; i <= m; i++) {
            int u, v, c; double p;
            scanf("%d%d%d%lf", &u, &v, &c, &p);
            add(u, v, 1, 0);
            if (c > 1) add(u, v, c - 1, -log(1.0 - p));
        }
        printf("%.2f\n", 1.0 - exp(-mcf(s, t)));
    }
    return 0;    
}

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[K. Finding Hotels]

KD-tree查最近点对。用最小花费和估价函数剪枝。

#include 
#define ll long long
#define pii pair
#define fi first
#define se second

namespace IO
{
    char buf[1 << 21], buf2[1 << 21], a[20], *p1 = buf, *p2 = buf, hh = ' ';
    int p, p3 = -1;
    void read() {}
    void print() {}
    inline int getc() {
        return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
    }
    inline void flush() {
        fwrite(buf2, 1, p3 + 1, stdout), p3 = -1;
    }
    template 
    inline void read(T &x, T2 &... oth) {
        T f = 1; x = 0;
        char ch = getc();
        while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getc(); }
        while (isdigit(ch)) { x = x * 10 + ch - 48; ch = getc(); }
        x *= f;
        read(oth...);
    }
    template 
    inline void print(T x, T2... oth) {
        if (p3 > 1 << 20) flush();
        if (x < 0) buf2[++p3] = 45, x = -x;
        do {
            a[++p] = x % 10 + 48;
        } while (x /= 10);
        do {
            buf2[++p3] = a[p];
        } while (--p);
        buf2[++p3] = hh;
        print(oth...);
    }
} // using namespace IO
#define read IO::read
#define print IO::print
#define flush IO::flush

template<class T> inline void checkmax(T &a, T b) { if (a < b) a = b; }
template<class T> inline void checkmin(T &a, T b) { if (a > b) a = b; }
ll sqr(int x) { return 1LL * x * x; }

const int N = 2e5 + 7;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
int n, m, D, root;
std::pii ans;

struct Node {
    int lp, rp, id;
    int d[3], mx[3], mn[3];
    inline bool operator < (const Node &rhs) const {
        return d[D] < rhs.d[D];
    }
    void clear() {
        lp = rp = id = 0;
        for (int i = 0; i < 3; i++)
            d[i] = mx[i] = mn[i] = 0;
    }
} tree[N], qu, in[N];

struct Kd {
    #define lp tree[p].lp
    #define rp tree[p].rp
    inline void pushup(int p, int s) {
        for (int i = 0; i < 3; i++)
            checkmax(tree[p].mx[i], tree[s].mx[i]),
            checkmin(tree[p].mn[i], tree[s].mn[i]);
    }
    inline void pushup(int p) {
        for (int i = 0; i < 3; i++)
            tree[p].mx[i] = tree[p].mn[i] = tree[p].d[i];
        if (lp) pushup(p, lp);
        if (rp) pushup(p, rp);
    }
    int build(int l, int r, int d) {
        D = d;
        int mid = l + r >> 1, p = mid;
        std::nth_element(tree + l, tree + mid, tree + r + 1);
        if (l != mid) lp = build(l, mid - 1, (d + 1) % 3);
        else lp = 0;
        if (r != mid) rp = build(mid + 1, r, (d + 1) % 3);
        else rp = 0;
        pushup(p);
        return p;
    }
    ll gu(int p) {
        if (qu.d[2] < tree[p].mn[2]) return inf + 1;
        ll ret = 0;
        for (int i = 0; i < 2; i++) {
            if (qu.d[i] > tree[p].mx[i]) ret += sqr(qu.d[i] - tree[p].mx[i]);
            if (qu.d[i] < tree[p].mn[i]) ret += sqr(qu.d[i] - tree[p].mn[i]);
        }
        return ret;
    }
    ll dis(int p) {
        if (qu.d[2] < tree[p].d[2]) return inf + 1;
        ll ret = 0;
        for (int i = 0; i < 2; i++)
            ret += sqr(qu.d[i] - tree[p].d[i]);
        return ret;
    }
    void query(int p) {
        if (!p) return;
        ll cur = dis(p);
        if (cur < ans.fi || (cur == ans.fi && tree[p].id < ans.se))
            ans.fi = cur, ans.se = tree[p].id;
        ll dl = 0, dr = 0;
        if (lp) dl = gu(lp);
        if (rp) dr = gu(rp);
        if (dl < dr) {
            if (dl <= ans.fi && lp) query(lp);
            if (dr <= ans.fi && rp) query(rp);
        } else {
            if (dr <= ans.fi && rp) query(rp);
            if (dl <= ans.fi && lp) query(lp);
        }
    }
} kd;

int main() {
    freopen("in.txt", "r", stdin);
    int T;
    read(T);
    for ( ; T--; ) {
        read(n, m);
        for (int i = 1; i <= n; i++) {
            tree[i].clear();
            tree[i].id = i;
            for (int j = 0; j < 3; j++)
                read(tree[i].d[j]);
            in[i] = tree[i];
        }
        root = kd.build(1, n, 1);
        for (int i = 1; i <= m; i++) {
            ans.fi = inf, ans.se = 0;
            for (int j = 0; j < 3; j++)
                read(qu.d[j]);
            kd.query(root);
            printf("%d %d %d\n", in[ans.se].d[0], in[ans.se].d[1], in[ans.se].d[2]);
        }
    }
    return 0;
}

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