Codeforces Round #420 (Div. 2)

出锅很严重…最后unrated…

B. Okabe and Banana Trees

大力枚举

C. Okabe and Boxes

题意：给定一个栈的2N种操作，任意两次操作之间可以用1的代价重构栈，问最小代价使得出栈序列为1,2,3,...,n.

考虑维护一个堆和一个栈。加入元素时放到栈顶，删除时分三类讨论：

1. 栈为空，弹出堆顶元素
2. 栈顶为最小元素，弹出
3. 否则将栈中元素全部入堆，栈清空，弹出堆顶。

正确性显然，每个元素入栈出栈入堆出堆各一次，复杂度O(nlgn)

#include <bits/stdc++.h>using namespace std;priority_queue<int, vector<int>, greater<int> > que;stack<int> stk;int n;char str[20];int dt;int main(){    scanf("%d", &n);    int k = 0, cnt = 0;    for (int i = 1; i <= n*2; i++) {        scanf("%s", str);        if (str[0] == 'r') {            k++;            if (stk.empty()) {                que.pop();            } else if (stk.top() == k){                stk.pop();            } else {                cnt++;                while (!stk.empty())                    que.push(stk.top()), stk.pop();                que.pop();            }        } else {            scanf("%d", &dt);            stk.push(dt);        }    }    cout << cnt << endl;    return 0;}

E. Okabe and El Psy Kongroo

分段dp，矩乘优化..太套路了对中国选手毫无杀伤力..

#include <bits/stdc++.h>using namespace std;int n;long long k;struct seg {    long long a, b, c;    friend bool operator < (const seg &a, const seg &b)    { return a.a < b.a;  }} s[105];const int N = 15, mod = 1e9+7;struct matrix {    int A[20][20];    matrix()    { memset(A, 0, sizeof A); }    friend matrix operator * (const matrix &a ,const matrix &b)    {        matrix c;        for (int i = 0; i <= N; i++)            for (int j = 0; j <= N; j++)                for (int k = 0; k <= N; k++)                    c.A[i][j] = (c.A[i][j]+(long long)a.A[i][k]*b.A[k][j])%mod;        return c;    }};matrix I;matrix power(matrix a, long long n){    matrix ans = I;    for (register int i = 0; i <= 61; i++) {        if ((n>>i)&1) ans = ans*a;        a = a*a;    }    return ans;}int main(){    scanf("%d%lld", &n, &k);    for (int i = 0; i <= N; i++) I.A[i][i] = 1;    for (int i = 1; i <= n; i++) {        scanf("%lld%lld%lld", &s[i].a, &s[i].b, &s[i].c);        if (s[i].b > k) s[i].b = k;    }    sort(s+1, s+n+1);    for (int i = 1; i < n; i++) {        if (s[i].c <= s[i+1].c) s[i+1].a++;        else s[i].b--;    }    matrix A = I;    int last_add = 0;    for (int i = 1; i <= n; i++) {            matrix C;        if (s[i].a > s[i].b) continue;        for (int j = 0; j <= s[i].c; j++) {            if (j-1 >= 0) C.A[j][j-1]++;            C.A[j][j]++;            if (j+1 <= N) C.A[j][j+1]++;        }        A = power(C, s[i].b-s[i].a+(i!=1))*A;    }    cout << A.A[0][0] << endl;    return 0;}