蓝桥杯2022扫雷_二分枚举

P2038 - [蓝桥杯2022初赛] 扫雷 - New Online Judgehttp://oj.ecustacm.cn/problem.php?id=2038

暴力能过30%

#include
using namespace std;
const int N = 5e4 + 4;
struct node {
    int x, y, r;
};
node a[N], b[N];
int vis[N] = {0};
int n, m, ans = 0;
int within(node p, node q) {
    int dis2 = (p.x - q.x) * (p.x - q.x) + (p.y - q.y) * (p.y - q.y);
    return dis2 <= q.r * q.r;
}
void blowUp(node q) {
    for (int i = 0; i < n; i++) {
        if (!vis[i]) {
            if (within(a[i], q)) {
                vis[i] = 1;
                ans++;
                blowUp(a[i]);
            }
        }
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        cin >> a[i].x >> a[i].y >> a[i].r;
    }
    for (int i = 0; i < m; i++) {
        cin >> b[i].x >> b[i].y >> b[i].r;
    }
    for (int j = 0; j < m; j++) {
        blowUp(b[j]);
    }
    cout << ans;
    return 0;
}

AC代码：

#include 
using namespace std;
const int N = 5e4 + 4;
struct node {
    int x, y, r;
};
node a[N], b[N];
int vis[N] = {0};
int n, m, ans = 0;
int within(node& p, node& q) {
    int dis2 = (p.x - q.x) * (p.x - q.x) + (p.y - q.y) * (p.y - q.y);
    return dis2 <= q.r * q.r;
}
int first(node& q) {
    int low = 0, high = n - 1;
    while (low < high) {
        int mid = (low + high) >> 1;
        if (a[mid].x < q.x - q.r) {
            low = mid + 1;
        } else {
            high = mid;
        }
    }
    return low;
}
void blowUp(node& q) {
    int i = first(q);
    for (; i < n; i++) {
        if (a[i].x > q.x + q.r) {
            break;
        }
        if (!vis[i]) {
            if (within(a[i], q)) {
                vis[i] = 1;
                ans++;
                blowUp(a[i]);
            }
        }
    }
}
int cmp(node& p, node& q) {
    return p.x < q.x;
}
int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        cin >> a[i].x >> a[i].y >> a[i].r;
    }
    for (int i = 0; i < m; i++) {
        cin >> b[i].x >> b[i].y >> b[i].r;
    }
    sort(a, a + n, cmp);
    for (int j = 0; j < m; j++) {
        blowUp(b[j]);
    }
    cout << ans;
    return 0;
}

先对炸雷按x的值排序，每次爆炸的q能引爆的炸雷的x值肯定在区间[q.x-q.r,q.x+q.r]内，所以可以二分枚举出第一个x值>=q.x-q.r的炸雷，减少每次遍历的范围