[代码]ACM-ICPC 2012 Regionals Asia - Changchun A Alice and Bob / ZJU 3655

Abstract

ZJU 3655 Alice and Bob

乱搞 离散化 implementation

 

Body

Source

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3655

Description

才不告诉你呢自己看去>_<

Solution

显然，Alice最后可以拿到a-b个。考虑对于Alice的某个决策区间（长度为a），Bob一定会选择其中一个长度为a-b的区间，使得区间内的白球个数最少。

所以这实质上就是个RMQ问题找区间最小值。由于需要处理的区间很多，所以需要离散化，只留下那些本质不同的区间。

至于离散化的过程就是恶心又tricky的implementation……

Code

即使你注意到R-L+1爆long long而启用unsigned long long，也还是要各种小心，少用减法，步步为营……

我是4Y的……给1Y的bluemary跪了……

#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

#include <algorithm>

using namespace std;



#define fr first

#define sc second

#define L(x) (x<<1)

#define R(x) (x<<1|1)

#define M(x) (x>>1)



typedef unsigned long long ll;

typedef pair<ll, int> pr;

pr (*mkpr)(ll, int) = make_pair<ll, int>;



int N, M;

ll l, r, a, b, c;

ll p[111111];



deque<ll> q;

vector<pr> node;



struct st {

    int l, r, v;

}t[222222<<2];



void pushup(int n) {

    t[n].v = min(t[L(n)].v, t[R(n)].v);

}



void build(int l, int r, int n) {

    t[n].l = l, t[n].r = r;

    if (l==r) {

        t[n].v = node[l].sc;

        return;

    }

    int m = M(l+r);

    build(l, m, L(n));

    build(m+1, r, R(n));

    pushup(n);

}



int query(int l, int r, int n) {

    if (l<=t[n].l && r>=t[n].r) return t[n].v;

    int res = 0x3fffffff;

    int m = M(t[n].l+t[n].r);

    if (l<=m) res = min(res, query(l, r, L(n)));

    if (r>m) res = min(res, query(l, r, R(n)));

    return res;

}



int bs(ll x) {

    int l = 0, r = M-1, m;

    while (l<r) {

        m = l+r+1>>1;

        if (node[m].fr<=x) l = m;

        else r = m-1;

    }

    return l;

}



int main() {

    int i, j, k;

    ll now;

    while (cin>>N) {

        cin>>l>>r>>a>>b;

        c = a-b;

        for (i = 0; i < N; ++i)

            scanf("%llu", p+i);

        if (a==b) {

            puts("0");

            continue;

        }

        i = 0;

        now = l+c-1;

        node.clear();

        q.clear();

        while (now <= r) {

            while (i<N && p[i]<=now) q.push_back(p[i++]);

            while (!q.empty() && q.front()+c<=now) q.pop_front();

            node.push_back(mkpr(now, (int)q.size()));

            if (i==N && q.empty()) break;

            ll add = i<N ? p[i]-now : INF;

            ll del = q.empty() ? INF : q.front()+c-now;

            now += min(add, del);

        }

        if (node.size()==0) {

            puts("0");

            continue;

        }

        M = node.size();

        build(0, M-1, 1);

        int ans = 0;

        for (i = 0; i < M; ++i) {

            ll up = node[i].fr+b;

            if (up>r) continue;

            j = bs(up);

            if (i<=j) ans = max(ans, query(i, j, 1));

        }

        for (i = 0; i < M; ++i) {

            if (node[i].fr+1<l+a) continue;

            ll down = node[i].fr-b;

            j = bs(down);

            if (j<=i) ans = max(ans, query(j, i, 1));

        }

        printf("%d\n", ans);

    }

    return 0;

}