UVA1492 - Adding New Machine(扫描线)

UVA1492 - Adding New Machine(扫描线)

题目链接

题目大意：给你N∗M个格子，这些格子中某些格子是放了旧的机器，然后问现在要在这些格子放一台1∗M的新机器，问有多少种放法。

解题思路：这题照样是可以转换成面积并来做，对于有旧机器（x，y）的格子，那么(x - M + 1,y)都是不可以放新机器的格子，还有从(H - M + 2,H)都是不可以放新机器的格子，所以覆盖的范围就要扩大。用扫描线算出这些不可以放新机器的格子，然后用总共的格子数剪掉就得到答案。分横着放和竖着放两种情况。注意M = 1的时候要特判，因为不存在横着和竖着两种情况。

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 5e4 + 5;
typedef long long ll;
#define lson(x) (x<<1)
#define rson(x) ((x<<1) | 1)

int x[2][maxn], y[2][maxn];

struct Node {

    int l, r, add, s;
    void set (int l, int r, int add, int s) {

        this->l = l;
        this->r = r;
        this->add = add;
        this->s = s;
    }
}node[8 * maxn];

struct Line {

    int x, y1, y2, flag;
    Line (int x, int y1, int y2, int flag) {

        this->x = x;
        this->y1 = y1;
        this->y2 = y2;
        this->flag = flag;
    }

    bool operator < (const Line& a) const {
        return x < a.x;
    }
};

vector<int> pos;
vector<Line> L;
int W, H, N, M;    

void pushup (int u) {

    if (node[u].add)
        node[u].s = pos[node[u].r + 1] - pos[node[u].l];
    else if (node[u].l == node[u].r) 
        node[u].s = 0;
    else
        node[u].s = node[lson(u)].s + node[rson(u)].s;
}

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

    node[u].set (l, r, 0, 0);
    if (l == r)
        return;

    int m = (l + r)>>1;
    build (lson(u), l, m);
    build (rson(u), m + 1, r);
    pushup(u);
}

void update (int u, int l, int r, int v) {

    if (node[u].l >= l && node[u].r <= r) {

        node[u].add += v;
        pushup(u);
        return ;
    }

    int m = (node[u].l + node[u].r)>>1;
    if (l <= m)
        update (lson(u), l, r, v);
    if (r > m)
        update (rson(u), l, r, v);
    pushup(u);
}

void init () {

    for (int i = 0; i < N; i++) 
        scanf ("%d%d%d%d", &x[0][i], &y[0][i], &x[1][i], &y[1][i]);        
}

ll solve (int w, int h, int x[2][maxn], int y[2][maxn]) {

    L.clear();
    pos.clear();
    int tmp;

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

        tmp = max(y[0][i] - M + 1, 1);    
        L.push_back(Line(x[0][i], tmp, y[1][i] + 1, 1));    
        L.push_back(Line(x[1][i] + 1, tmp, y[1][i] + 1, -1));
        pos.push_back(tmp);
        pos.push_back(y[1][i] + 1);
    }

    tmp = max(1, h - M + 2);
    L.push_back(Line(1, tmp, h + 1, 1));
    L.push_back(Line(w + 1, tmp, h + 1, -1));
    pos.push_back(tmp);
    pos.push_back(h + 1);

    sort (L.begin(), L.end());
    sort (pos.begin(), pos.end());
    pos.erase (unique(pos.begin(), pos.end()), pos.end());

    build(1, 0, (int)pos.size() - 1);

    ll ans = 0;
    int l, r;
    for (int i = 0; i < L.size() - 1; i++)  {

        l = lower_bound(pos.begin(), pos.end(), L[i].y1) - pos.begin();
        r = lower_bound(pos.begin(), pos.end(), L[i].y2) - pos.begin();
        update(1, l, r - 1, L[i].flag);
// printf ("%d %d\n", node[1].s, L[i + 1].x - L[i].x);
        ans += (ll)node[1].s * (L[i + 1].x - L[i].x);
    }

    return ans;
}

int main () {

    ll ans;
    while (scanf ("%d%d%d%d", &W, &H, &N, &M) != EOF) {

        init();

        if (M == 1) {
            ans = 0;
            for (int i = 0; i < N; i++)
                ans += (ll) (x[1][i] + 1 - x[0][i]) * (y[1][i] + 1- y[0][i]);    
            ans = (ll)W * H - ans;
        } else 
            ans = 2 * (ll)W * H - solve(H, W, y, x) - solve(W, H, x, y);
        printf ("%lld\n", ans);
    }
    return 0;
}