暑假集训-数据结构

求矩形的并 Poj1151

线段树，离散化，扫描线。

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define LL long long
#define INF 0x3f3f3f3f
#define OPEN_FILE
using namespace std;
const int MAXN = 5005;
struct Line{
    double left, right;
    double y;
    int cnt;
    Line(){};
    Line(double left, double right, double y, int cnt) :left(left), right(right), y(y), cnt(cnt){};
};
bool operator < (Line a, Line b){
    return a.y < b.y;
}
struct Node{
    double left, right;
    int from, to;
    int cnt;
    double value;
};
int n;
double a[MAXN];
Line scan[MAXN];
Node f[MAXN << 1];
void build(int t, int left, int right){
    f[t].left = a[left];
    f[t].right = a[right];
    f[t].from = left;
    f[t].to = right;
    f[t].cnt = f[t].value = 0;
    if (right - left == 1) return;
    int mid = (left + right) >> 1;
    build(t << 1, left, mid);
    build((t << 1) + 1, mid, right);
}
void update(int t, Line p){
    if (p.left == f[t].left && p.right == f[t].right){
        f[t].cnt += p.cnt;
        if (f[t].cnt > 0){
            f[t].value = p.right - p.left;
            return;
        }
        if (f[t].to - f[t].from == 1){
            f[t].value = 0;
        }
        else{
            f[t].value = f[t << 1].value + f[(t << 1) + 1].value;
        }
        return;
    }
    if (p.left >= f[(t << 1) + 1].left){
        update((t << 1) + 1, p);
    }
    else if (p.right <= f[t << 1].right){
        update(t << 1, p);
    }
    else{
        Line q = p;
        q.right = f[t << 1].right;
        update(t << 1, q);
        q = p;
        q.left = f[(t << 1) + 1].left;
        update((t << 1) + 1, q);
    }
    if (f[t].cnt > 0){
        f[t].value = f[t].right - f[t].left;
    }
    else{
        f[t].value = f[t].to - f[t].from == 1 ?0:f[t << 1].value + f[(t << 1) + 1].value;
    }
}
int main()
{
#ifdef OPEN_FILE
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // OPEN_FILE
    double x1, y1, x2, y2;
    int cas = 1;
    while (~scanf("%d", &n) && n != 0){
        for (int i = 1; i <= n; i++){
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            a[i * 2 - 1] = x1;
            a[i * 2] = x2;
            scan[i * 2 - 1] = Line(x1, x2, y1, 1);
            scan[i * 2] = Line(x1, x2, y2, -1);
        }
        n = n * 2;
        double ans = 0.0;
        memset(f, 0, sizeof(f));
        sort(a + 1, a + n + 1);
        sort(scan + 1, scan + n + 1);
        build(1, 1, n);
        update(1, scan[1]);
        for (int i = 2; i <= n; i++){
            ans += f[1].value * (scan[i].y - scan[i - 1].y);
            update(1, scan[i]);
        }
        printf("Test case #%d
Total explored area: %.2lf

", cas++, ans);
    }
}

并查集应用 poj1182

以前对并查集的印象还停留在高中，只是记录一下是不是已经连通。这里加了一个数组rank[i]表示i的秩，也就是到祖先的距离，其实都是高中的内容。

//set
void MakeSet(int x){
    father[x] = x;
    rank[x] = 0;
}
//find
int FindSet(int x)
{
    if (x != father[x])
    {
        father[x] = FindSet(father[x]);
    }
    return father[x];
}
//union
void Union(int x, int y)
{
    x = FindSet(x);
    y = FindSet(y);
    if (x == y) return;
    if (rank[x] > rank[y])
    {
        father[y] = x;
        rank[x] += rank[y];
    }
    else
    {
        if (rank[x] == rank[y])
        {
            rank[y]++;
        }
        father[x] = y;
    }
}