BZOJ 2716: [Violet 3]天使玩偶

2716: [Violet 3]天使玩偶

Time Limit: 80 Sec  Memory Limit: 128 MB
Submit: 1473  Solved: 621
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Description

Input

Output

Sample Input & Output

样例过大，略

HINT

Source

Vani原创 欢迎移步 OJ2648

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CDQ分治，分类讨论拆绝对值的方式，分别查询最优值。

#include <cstdio>
#include <cstring>
#include <algorithm>

inline int nextChar(void) {
    const int siz = 1024;
    static char buf[siz];
    static char *hd = buf + siz;
    static char *tl = buf + siz;
    if (hd == tl)
        fread(hd = buf, 1, siz, stdin);
    return int(*hd++);
}

inline int nextInt(void) {
    register int ret = 0;
    register int neg = false;
    register int bit = nextChar();
    for (; bit < 48; bit = nextChar())
        if (bit == '-')neg ^= true;
    for (; bit > 47; bit = nextChar())
        ret = ret * 10 + bit - 48;
    return neg ? -ret : ret;
}

const int lim = 1000001;
const int siz = 1000005;
const int inf = 1000000007;

int n, m;

struct data {
    int k, x, y, t, p;
    inline friend bool operator < 
    (const data &a, const data &b) {
        if (a.x != b.x)
            return a.x < b.x;
        else
            return a.k < b.k;
    }
}p[siz], s[siz], q[siz];

int ans[siz];

int now;
int bit[siz];
int tim[siz];

inline void add(int t, int k) {
    for (; t < siz; t += t&-t)
        if (bit[t] < k || tim[t] != now)
            bit[t] = k, tim[t] = now;
}

inline int ask(int t) {
    int ret = -inf;
    for (; t; t -= t&-t)
        if (ret < bit[t] && tim[t] == now)
            ret = bit[t];
    return ret;
}

void cdqSolve(int l, int r) {
    if (l >= r)return;
    
    int mid = (l + r) >> 1;
    
    cdqSolve(l, mid);
    cdqSolve(mid + 1, r);
    
    int t1 = l, t2 = mid + 1, tot = l;
    
    while (t1 <= mid && t2 <= r) {
        if (s[t1] < s[t2])
            q[tot++] = s[t1++];
        else
            q[tot++] = s[t2++];
    }
    
    while (t1 <= mid)
        q[tot++] = s[t1++];
        
    while (t2 <= r)
        q[tot++] = s[t2++];
        
    for (int i = l; i <= r; ++i)
        s[i] = q[i];
    
    ++now;
    
    for (int i = l; i <= r; ++i)
        if (s[i].k && s[i].t > mid) {
            int tmp = s[i].x + s[i].y - ask(s[i].y);
            if (ans[s[i].p] > tmp)
                ans[s[i].p] = tmp;
        }
        else if (!s[i].k && s[i].t <= mid)
            add(s[i].y, s[i].x + s[i].y);
}

inline void cdqSolve1(void) {
    memcpy(s, p, sizeof(s));
    cdqSolve(1, n + m);
}

inline void cdqSolve2(void) {
    memcpy(s, p, sizeof(s));
    for (int i = 1; i <= n + m; ++i)
        s[i].x = lim - s[i].x;
    cdqSolve(1, n + m);
}

inline void cdqSolve3(void) {
    memcpy(s, p, sizeof(s));
    for (int i = 1; i <= n + m; ++i)
        s[i].y = lim - s[i].y;
    cdqSolve(1, n + m);
}

inline void cdqSolve4(void) {
    memcpy(s, p, sizeof(s));
    for (int i = 1; i <= n + m; ++i)
        s[i].x = lim - s[i].x,
        s[i].y = lim - s[i].y;
    cdqSolve(1, n + m);
}

signed main(void) {
//    freopen("in", "r", stdin);
//    freopen("out", "w", stdout);
    
    n = nextInt();
    m = nextInt();
    
    for (int i = 1; i <= n; ++i) {
        p[i].x = nextInt();
        p[i].y = nextInt();
        p[i].k = 0;
        p[i].t = i;
    }
    
    for (int i = 1; i <= m; ++i) {
        p[i + n].k = nextInt() - 1;
        p[i + n].x = nextInt();
        p[i + n].y = nextInt();
        p[i + n].t = i + n;
        p[i + n].p = i;
    }
    
    for (int i = 1; i <= m; ++i)
        ans[i] = inf;
    
    cdqSolve1();
    cdqSolve2();
    cdqSolve3();
    cdqSolve4();
    
    for (int i = 1; i <= m; ++i)
        if (p[i + n].k)printf("%d
", ans[i]);
}

@Author: YouSiki