BZOJ3476 [Usaco2014 Mar]The Lazy Cow

这题做的我真是23333。。。

奥，这题表述不清，是说曼哈顿距离 <= k

首先我们坐标变换：

设(X, Y)为原坐标，则新坐标为(X + Y, X - Y + E)，为了防止y < 0

然后我们只要维护一堆边平行于坐标轴的正方形即可

方法是扫描线扫过去，然后用线段树维护y轴上的权值，也就是维护段+1和求全部的最大值操作。

然后发现Rank2。。。为了Rank1开始丧心病狂，写起了入读优化、、、

发现差4ms就Rank1，极为不爽。。。

于是czr大爷告诉我可以fread。。。一番折腾后。。。

Rank1！欧液！

/**************************************************************
    Problem: 3476
    User: rausen
    Language: C++
    Result: Accepted
    Time:2192 ms
    Memory:64480 kb
****************************************************************/
 
#include <cstdio>
#include <cctype>
#include <algorithm>
 
#define lson p << 1
#define rson p << 1 | 1
#define Tag seg[p].tag
using namespace std;
const int N = 100005;
const int E = 1000001;
const int Height = (E << 1) - 1;
const int M = (E << 3) + 10;
 
struct point {
    int x, y, v;
}a[N];
inline bool operator < (const point &a, const point &b) {
    return a.x < b.x;
}
 
struct segment_tree {
    int tag, v;
}seg[M];
 
int n, k;
int L, R, del;
 
inline unsigned int read() {
    int x = 0;
    char ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}
 
inline void refresh(int p, int v) {
    seg[p].v += v;
    seg[p].tag += v;
}
 
void insert(int p, int l, int r) {
    if (L <= l && r <= R) {
        refresh(p, del);
        return;
    }
    if (Tag) {
        refresh(lson ,Tag), refresh(rson, Tag);
        Tag = 0;
    }
    int mid = l + r >> 1;
    if (L <= mid)
        insert(lson, l, mid);
    if (mid < R) 
        insert(rson, mid + 1, r);
    seg[p].v = max(seg[lson].v, seg[rson].v);
}
 
int main() {    
    int i, j, X, Y, ans = 0;
    n = read(), k = read();
    k = k * 2 + 1;
    for (i = 1; i <= n; ++i) {
        a[i].v = read(), X = read(), Y = read();
        a[i].x = X + Y, a[i].y = X - Y + E;
    }
    sort(a + 1, a + n + 1);
    for (i = j = 1; i <= n; ++i) {
        while (a[i].x - a[j].x + 1 > k) {
            L = a[j].y - k + 1, R = a[j].y, del = -a[j].v;
            insert(1, 1, Height), ++j;
        }
        L = a[i].y - k + 1, R = a[i].y, del = a[i].v;
        insert(1, 1, Height);
        ans = max(ans, seg[1].v);
    }
    printf("%d
", ans);
    return 0;
}

↓下面这就是那个坑爹fread的版本。。。：

/**************************************************************
    Problem: 3476
    User: rausen
    Language: C++
    Result: Accepted
    Time:1940 ms
    Memory:66336 kb
****************************************************************/
 
#include <cstdio>
#include <cctype>
#include <algorithm>
 
#define lson p << 1
#define rson p << 1 | 1
#define Tag seg[p].tag
using namespace std;
const int N = 100005;
const int E = 1000001;
const int Height = (E << 1) - 1;
const int M = (E << 3) + 10;
const int MaxCH = 1900100;
struct point {
    int x, y, v;
}a[N];
inline bool operator < (const point &a, const point &b) {
    return a.x < b.x;
}
 
struct segment_tree {
    int tag, v;
}seg[M];
 
int n, k;
int L, R, del;
char CH[MaxCH];
int len, Left = 0;
 
inline unsigned int read() {
    int x = 0;
    char ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}
 
inline int find() {
    int x = 0;
    while (CH[Left] < '0' || '9' < CH[Left])
        ++ Left;
    while ('0' <= CH[Left] && CH[Left] <= '9')
        x = x * 10 + CH[Left++] - '0';
    return x;
}
 
inline void refresh(int p, int v) {
    seg[p].v += v;
    seg[p].tag += v;
}
 
void insert(int p, int l, int r) {
    if (L <= l && r <= R) {
        refresh(p, del);
        return;
    }
    if (Tag) {
        refresh(lson, Tag), refresh(rson, Tag);
        Tag = 0;
    }
    int mid = l + r >> 1;
    if (L <= mid)
        insert(lson, l, mid);
    if (mid < R) 
        insert(rson, mid + 1, r);
    seg[p].v = max(seg[lson].v, seg[rson].v);
}
 
int main() {    
    int i, j, X, Y, ans = 0;
    len = fread(CH, 1, MaxCH, stdin);
    CH[len] = ' ';
    n = find(), k = find();
    k = k * 2 + 1;
    for (i = 1; i <= n; ++i) {
        a[i].v = find(), X = find(), Y = find();
        a[i].x = X + Y, a[i].y = X - Y + E;
    }
    sort(a + 1, a + n + 1);
    for (i = j = 1; i <= n; ++i) {
        while (a[i].x - a[j].x + 1 > k) {
            L = a[j].y - k + 1, R = a[j].y, del = -a[j].v;
            insert(1, 1, Height), ++j;
        }
        L = a[i].y - k + 1, R = a[i].y, del = a[i].v;
        insert(1, 1, Height);
        ans = max(ans, seg[1].v);
    }
    printf("%d
", ans);
    return 0;
}