其他-学长杂题讲义

1. Rasheda And The Zeriba

大致题意： 给出n条边，判断能否构成一个n边凸多边形，并求出能覆盖该图形的圆的最小半径 1 ≤ n ≤ 1000 。

http://codeforces.com/gym/100283/problem/A 

#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

int N;
double R[1005];
const double _pi = asin(1)*2;

bool Check(double v)
{
    double tmp = 0; int i;
    for (i = 1; i <= N; ++i)
        if ((tmp += asin(R[i]/2/v)*2) > 2*_pi) break;
    return i == N+1;
}

int main()
{
    freopen("zeriba.in", "r", stdin);
    int T;
    scanf("%d", &T);
    for (int ca = 1; ca <= T; ++ca) {
        scanf("%d", &N);
        double sum = 0, mx = 0; int i;
        for (i = 1; i <= N; ++i)
            scanf("%lf", &R[i]), sum += R[i], mx = max(mx, R[i]);
        for (i = 1; i <= N; ++i)
            if (sum-R[i] <= R[i]) break;
        if (i != N+1) { printf("Case %d: can't form a convex polygon
", ca); continue; }
        double l = mx/2, r = sum/4;
        while (fabs(l-r) >= 0.000001) {
            double mid = (l+r)/2;
            if (Check(mid)) r = mid;
            else l = mid;
        }
        printf("Case %d: %.4lf
", ca, l);
    }
    return 0;
}

 2. Fitting boxes

大致题意：给出两个矩形长宽（1 <= x, y, a, b <= 1000），判断能否使其中一个完全覆盖另一个。

http://codeforces.com/gym/100738/problem/A

#include <cstdio>
#include <cmath>
#include <algorithm>

const double _PI = asin(1)*2;

using namespace std;

inline double P(double v) { return v * v; }

bool Test(double x, double y, double a, double b)
{
    if (x*y < a*b || a > x) return false;
    if (b <= y) return true;
    if (P(y*b) < P(a*a)+P(a*b)-P(a*y)) return false;
    
    double tmp = _PI/2 - (atan(sqrt(P(y)/(P(a)+P(b)-P(y)))) - atan(a/b));
    double tmp2 = b*sin(tmp)+a*cos(tmp);
    return 0 <= tmp2 && tmp2 <= x;
}

int main()
{
    double t1, t2, t3, t4;
    int ans = 0;
    scanf("%lf%lf%lf%lf", &t1, &t2, &t3, &t4);

    ans |= Test(t1, t2, t3, t4);
    ans |= Test(t1, t2, t4, t3);
    ans |= Test(t2, t1, t3, t4);
    ans |= Test(t2, t1, t4, t3);
    
    swap(t1, t3), swap(t2, t4);
    
    ans |= Test(t1, t2, t3, t4);
    ans |= Test(t1, t2, t4, t3);
    ans |= Test(t2, t1, t3, t4);
    ans |= Test(t2, t1, t4, t3);
    
    if (ans) printf("Yes
");
    else printf("No
");
    return 0;
}

3. Pretty Buses

大致题意：一个n个点m条边的无向图，每条边有一个权值。初始时，点i处有巴士i和司机i，你可以有两种操作:(1)Drive b x y,把巴士b从x点开到y点并付出这条边的权值的费用；(2)Move d x y，司机d从巴士x移动的巴士y上，没有花费 所有司机要集中在一个点，并且每个司机的换乘次数不多于25 求最小花费，并给出一种合理的方案 1 ≤ n ≤ 200000, 1 ≤ m ≤ 400000

http://codeforces.com/gym/100738/problem/E

#include <cstdio>
#include <queue>
#include <vector>

using namespace std;

const int _N = 400005;

struct tmpeg {
    int x, y, w;
    tmpeg(int x, int y, int w):
        x(x), y(y), w(w) { }
    bool operator < (const tmpeg t) const
    {
        return w > t.w;
    }
};

priority_queue<tmpeg> Q;
vector<int> G[_N], dri[_N];
int veh[_N], siz[_N], Dad[_N];

inline void Ins(int x, int y) { G[x].push_back(y); return; }

void diudiu(int node, int dad)
{
    int mx = node;
    for (int i = G[node].size()-1; i >= 0; --i) {
        int v = G[node][i];
        if (v == dad) continue;
        diudiu(v, node);
        printf("Drive %d %d %d
", veh[v], v, node);
        int a = veh[node], b = veh[v];
        if (siz[a] < siz[b]) swap(a, b);
        veh[node] = veh[v] = a;
        siz[a] += siz[b];
        for (int j = dri[b].size()-1; j >= 0; --j) {
            printf("Move %d %d %d
", dri[b][j], b, a);
            dri[a].push_back(dri[b][j]), veh[dri[b][j]] = a;
        }
        
    }
    return;
}

int GetDad(int v) { return Dad[v] == v ? v : Dad[v] = GetDad(Dad[v]); }

int main()
{
    int N, M, i;
    scanf("%d%d", &N, &M);
    for (i = 1; i <= M; ++i) {
        int x, y, w;
        scanf("%d%d%d", &x, &y, &w);
        Q.push(tmpeg(x, y, w));
    }
    long long ans = 0; int cnt = 0;
    for (i = 1; i <= N; ++i)
        Dad[i] = i, siz[i] = 1, veh[i] = i, dri[i].push_back(i);
    while (true) {
        tmpeg tmp = Q.top(); Q.pop();
        int tx = GetDad(tmp.x), ty = GetDad(tmp.y);
        if (tx != ty) {
            Dad[tx] = ty, Ins(tmp.x, tmp.y), Ins(tmp.y, tmp.x);
            ++cnt, ans += tmp.w;
            if (cnt == N-1) break;
        }
    }
    printf("%I64d
", ans);
    
    diudiu(1, 0);
    printf("Done
");
    return 0;
}

4. Mishka and Interesting sum

大致题意：询问数列 A 中 [L, R] 区间出现次数为偶数次的数字，每个数只算一次的异或和。比如对于 {1, 3, 3, 3, 4, 4} ，答案是 1^3^4 。

http://codeforces.com/contest/703/problem/D

#include <cstdio>
#include <algorithm>

using namespace std;

const int _N = 1050000;

struct query {
    int l, r, id;
    bool operator < (const query tmp) const
    {
        return this->l < tmp.l;
    }
} Q[_N];

int pos, N, M, N_A;
int A[_N], B[_N], C[_N], sum[_N], fst[_N], nxt[_N], ans[_N];
bool mk[_N];

void Add(int k, int d)
{
    for (int i = k; i <= N; i += i & -i)
        C[i] ^= d;
    return;
}

int GetSum(int k)
{
    int tmp = 0;
    for (int i = k; i; i ^= i & -i)
        tmp ^= C[i];
    return tmp;
}

int hs(int v) { return lower_bound(A+1, A+1+N_A, v)-A; }

int main()
{
    int i;
    scanf("%d", &N);
    for (i = 1; i <= N; ++i)
        scanf("%d", &A[i]), sum[i] = sum[i-1]^A[i], B[i] = A[i];
    scanf("%d", &M);
    for (i = 1; i <= M; ++i)
        scanf("%d%d", &Q[i].l, &Q[i].r), Q[i].id = i;
    
    sort(Q+1, Q+1+M);
    sort(A+1, A+1+N);
    N_A = unique(A+1, A+1+N)-A-1;
    for (i = 1; i <= N; ++i) {
        int tmp;
        if (!mk[tmp = hs(B[i])])
            mk[tmp] = true, Add(i, B[i]);
    }
    pos = 1;
    for (i = N; i >= 1; --i) {
        int tmp = hs(B[i]);
        nxt[i] = fst[tmp], fst[tmp] = i;
    }
    
    for (i = 1; i <= M; ++i) {
        while (pos < Q[i].l) {
            Add(pos, B[pos]);
            if (nxt[pos]) Add(nxt[pos], B[pos]);
            ++pos;
        }
        ans[Q[i].id] = GetSum(Q[i].r)^sum[Q[i].r]^sum[Q[i].l-1];
    }
    for (i = 1; i <= M; ++i)
        printf("%d
", ans[i]);
    return 0;
}

5. Queries

大致题意：维护一个序列，支持 1.单点修改 2.查询某区间 [l, r] 中所有合理的 [i, j] (l <= i <= j <= r) 的异或结果的和。用 Elf(i, j) 表示 [i, j] 中所有数的异或结果，则 [l, r] 查询的结果应为 ∑Elf(i, j)  ( [i, j] ⊆ [l, r] ) 。

http://codeforces.com/gym/100739/problem/A

一定注意，尽管根据定义 lsum0（表示左起前缀和中 0 的个数） = 区间长度 - lsum1 （表示左起前缀和中 1 的个数），但是最好在线段树中同时维护这两个变量，这样在查询操作时可以避免一些特判。理论上只记一个变量+特判是可以的，我最初也是这么写的，也就 Wa 了那么七八次吧，还要感谢 newuser 大佬指出我特判的愚蠢性……OrzOrzOrz

#include <cstdio>

const int MOD = 4001;
const int _N = 2800000;

int Tot, N, M;
int lsum0[_N], lsum[_N], rsum0[_N], rsum[_N], xos[_N], cnt[_N], lson[_N], rson[_N], root[11];

void Modify(int &p, int l, int r, int k, int v)
{
    if (!p) p = ++Tot;
    if (l == r) { lsum[p] = rsum[p] = cnt[p] = xos[p] = v, lsum0[p] = rsum0[p] = v^1; return; }
    int mid = l+r >> 1;
    if (!lson[p]) lson[p] = ++Tot;
    if (!rson[p]) rson[p] = ++Tot;
    if (k <= mid) Modify(lson[p], l, mid, k, v);
    else Modify(rson[p], mid+1, r, k, v);
    lsum[p] = (lsum[lson[p]] + (xos[lson[p]] ? lsum0[rson[p]] : lsum[rson[p]]))%MOD;
    rsum[p] = (rsum[rson[p]] + (xos[rson[p]] ? rsum0[lson[p]] : rsum[lson[p]]))%MOD;
    lsum0[p] = (lsum0[lson[p]] + (xos[lson[p]] ? lsum[rson[p]] : lsum0[rson[p]]))%MOD;
    rsum0[p] = (rsum0[rson[p]] + (xos[rson[p]] ? rsum[lson[p]] : rsum0[lson[p]]))%MOD;
    xos[p] = xos[lson[p]] ^ xos[rson[p]];
    cnt[p] = (cnt[lson[p]]+cnt[rson[p]])%MOD;
    cnt[p] = (cnt[p]+rsum[lson[p]]*lsum0[rson[p]]%MOD)%MOD;
    cnt[p] = (cnt[p]+rsum0[lson[p]]*lsum[rson[p]]%MOD)%MOD;
    return;
}

int Query(int &p, int l, int r, int s, int t, int &tmp_l, int &tmp_r, int &tmp_x, int &tmp_l0, int &tmp_r0)
{
    if (!p) p = ++Tot;
    if (s <= l && r <= t) { tmp_l = lsum[p], tmp_r = rsum[p], tmp_x = xos[p], tmp_l0 = lsum0[p], tmp_r0 = rsum0[p]; return cnt[p]; }
    int mid = l+r >> 1, l_l=0, l_r=0, l_x=0, r_l=0, r_r=0, r_x=0, l_cnt=0, r_cnt=0, l_l0=0, l_r0=0, r_l0=0, r_r0=0;
    if (t >= l && s <= mid) l_cnt = Query(lson[p], l, mid, s, t, l_l, l_r, l_x, l_l0, l_r0);
    if (t > mid && s <= r) r_cnt = Query(rson[p], mid+1, r, s, t, r_l, r_r, r_x, r_l0, r_r0);
    
    tmp_l = (l_l + (l_x ? r_l0 : r_l))%MOD;
    tmp_r = (r_r + (r_x ? l_r0 : l_r))%MOD;
    tmp_l0 = (l_l0 + (l_x ? r_l : r_l0))%MOD;
    tmp_r0 = (r_r0 + (r_x ? l_r : l_r0))%MOD;
    tmp_x = l_x ^ r_x;
    int tmp_cnt = (l_cnt+r_cnt)%MOD;
    tmp_cnt = (tmp_cnt+l_r*r_l0%MOD)%MOD;
    tmp_cnt = (tmp_cnt+l_r0*r_l%MOD)%MOD;
    return tmp_cnt;
}

int main()
{
    int i, j;
    scanf("%d%d", &N, &M);
    for (i = 1; i <= N; ++i) {
        int tmp;
        scanf("%d", &tmp);
        for (j = 0; j < 11; ++j)
            Modify(root[j], 1, N, i, (tmp>>j)&1);
    }
    while (M--) {
        int ins, x, y;
        scanf("%d%d%d", &ins, &x, &y);
        if (ins == 1) {//change
            for (j = 0; j < 11; ++j)
                Modify(root[j], 1, N, x, (y>>j)&1);
        } else {//query
            int ans = 0;
            int tl, tr, tx, tl0, tr0;
            for (j = 0; j < 11; ++j)
                ans = (ans+(Query(root[j], 1, N, x, y, tl, tr, tx, tl0, tr0)<<j)%MOD)%MOD;
            printf("%d
", (ans+MOD)%MOD);
        }
    }
    return 0;
}

6. Decomposition into Good Strings

大致题意：定义 good string 为正好包含 k 种不同字母的字符串（如当 k = 2 时，abbbba 是一个 good string）。给定 k 和一个长为 n （1 <= n <= 200000）的字符串，求该字符串的每个前缀最少可以被分割为多少个 good string 。即输出 n 个数，分别对应原字符串的 n 个前缀的答案。

http://codeforces.com/gym/100971/problem/M

#include <cstdio>

char A[250000];
int f[250000], mk[300], cnt, K, pos;

void Update(int i)
{
    if (++mk[A[i]] == 1) ++cnt;
    if (cnt < K) { f[i] = -1; return; }
    while (cnt > K) {
        if (!--mk[A[pos]]) --cnt;
        ++pos;
    }
    if (pos == -1) pos = 1;
    bool flag = false;
    while (f[pos-1] == -1) {
        if (mk[A[pos]] == 1) { flag = true; break; }
        --mk[A[pos]], ++pos;
    }
    if (flag) { f[i] = -1; return; }
    f[i] = f[pos-1]+1;
    return;
}

int main()
{
    char tt;
    int i;
    scanf("%d", &K);
    f[0] = 0, pos = -1;
    while ((tt = getchar()) < 'a' || tt > 'z');
        A[i = 1] = tt, Update(i);
    while ((tt = getchar()) >= 'a' && tt <= 'z')
        A[++i] = tt, Update(i);
    for (int t = 1; t <= i; ++t)
        printf("%d ", f[t]);
    return 0;
}