poj 2528 Mayor's posters 线段树 || 并查集 离线处理

题目链接

题意

用不同颜色的线段覆盖数轴，问最终数轴上有多少种颜色？

注：只有最上面的线段能够被看到；即，如果有一条线段被其他的线段给完全覆盖住，则这个颜色是看不到的。

法一：线段树

按题意按顺序模拟即可。

法二：线段树+离线

将整个过程倒过来看待，如果要加进去的线段所在的区域已经完全被覆盖，那么这条线段就没有贡献，否则就有\(1\)的贡献。

法三：并查集+离线

离线处理思想同上。

用\(fa[\ ]\)数组记录某个元素左边距其最近的没有被覆盖的点的坐标。那么对于当前覆盖的线段\([l,r]\)，只要\(find(r)\lt l\)，就意味着当前这一段已经完全被覆盖，所以贡献为\(0\).

并查集的思想类似BZOJ 3211 花神游历各国

注意点

这道题的坑点在于：离散化

如果普通地进行离散化，考虑

3
1 10
1 4
6 10

会被离散化成

3
1 4
1 2
3 4

本来有三种颜色，处理过之后就只有两种颜色了。

问题在于：顺序的连续并不代表位置的连续。

处理时在中间插入额外的点即可。

Code

Ver. 1

#include <stdio.h>
#include <algorithm>
#include <string.h>
#define lson rt<<1
#define rson rt<<1|1
#define maxn 100010
using namespace std;
typedef long long LL;
int a[maxn], l[maxn], r[maxn], b[maxn], ans;
bool vis[maxn];
struct tree { int l, r, c, flag; }tr[maxn*4];
void build(int rt, int l, int r) {
    tr[rt].l = l, tr[rt].r = r; tr[rt].flag = tr[rt].c = 0;
    if (l==r) return;
    int mid = l+r >> 1;
    build(lson, l, mid), build(rson, mid+1, r);
}
void push_down(int rt) {
    if (tr[rt].flag) {
        tr[lson].flag = tr[rson].flag
            = tr[lson].c = tr[rson].c = tr[rt].flag;
        tr[rt].flag = 0;
    }
}
void push_up(int rt) {
    tr[rt].c = tr[lson].c == tr[rson].c ? tr[lson].c : 0;
}
void modify(int rt, int l, int r, int c) {
    if (tr[rt].l == l && tr[rt].r == r) {
        tr[rt].c = tr[rt].flag = c;
        return;
    }
    push_down(rt);
    int mid = tr[rt].l + tr[rt].r >> 1;
    if (r <= mid) modify(lson, l, r, c);
    else if (l > mid) modify(rson, l, r, c);
    else modify(lson, l, mid, c), modify(rson, mid+1, r, c);
    push_up(rt);
}
void query(int rt) {
    if (tr[rt].l == tr[rt].r || tr[rt].c) {
        if (!vis[tr[rt].c] && tr[rt].c) ++ans, vis[tr[rt].c] = true;
        return;
    }
    push_down(rt);
    query(lson); query(rson);
}
void work() {
    int n;
    scanf("%d", &n);
    int tot=0;
    for (int i = 1; i <= n; ++i) {
        scanf("%d%d", &l[i], &r[i]);
        a[++tot] = l[i], a[++tot] = r[i];
    }
    sort(a+1,a+tot+1);
    tot = unique(a+1,a+tot+1)-a-1;

    b[1] = a[1]; int cnt = 1;
    for (int i = 2; i <= tot; ++i) {
        if (a[i]-a[i-1] == 1) b[++cnt] = a[i];
        else b[++cnt] = a[i-1] + 1, b[++cnt] = a[i];
    }

    build(1,1,cnt);
    for (int i = 1; i <= n; ++i) {
        int pl = lower_bound(b+1, b+1+cnt, l[i]) - b,
            pr = lower_bound(b+1, b+1+cnt, r[i]) - b;
        modify(1, pl, pr, i);
    }

    ans = 0; memset(vis, 0, sizeof(vis));
    query(1);
    printf("%d\n", ans);
}
int main() {
    int T;
    scanf("%d", &T);
    while (T--) work();
    return 0;
}

Ver. 2

#include <stdio.h>
#include <algorithm>
#include <map>
#define lson rt<<1
#define rson rt<<1|1
#define maxn 100010
using namespace std;
typedef long long LL;
int a[maxn], l[maxn], r[maxn], b[maxn];
struct tree { int l, r; bool cov, flag; }tr[maxn*4];
map<int,int> mp;
void build(int rt, int l, int r) {
    tr[rt].l = l, tr[rt].r = r; tr[rt].cov = 0; tr[rt].flag = 0;
    if (l==r) return;
    int mid = l+r >> 1;
    build(lson, l, mid), build(rson, mid+1, r);
}
void push_down(int rt) {
    if (tr[rt].flag) {
        tr[lson].cov = tr[rson].cov = tr[lson].flag = tr[rson].flag = 1;
        tr[rt].flag = 0;
    }
}
void push_up(int rt) {
    tr[rt].cov = tr[lson].cov && tr[rson].cov;
}
bool insert(int rt, int l, int r) {
    if (tr[rt].l==l && tr[rt].r==r) {
        if (tr[rt].cov) return true;
        tr[rt].cov = tr[rt].flag = true;
        return false;
    }
    push_down(rt);
    int mid = tr[rt].l + tr[rt].r >> 1;
    bool ret;
    if (r <= mid) ret = insert(lson, l, r);
    else if (l > mid) ret = insert(rson, l, r);
    else {
        //    注意！这里不能直接写成 ret = insert(lson, l, mid) & ans2 = insert(rson, mid+1, r);
        //    因为根据短路原理，如果前面的为假，后面的就直接不做了！
        bool ans1 = insert(lson, l, mid), ans2 = insert(rson, mid+1, r);
        ret = ans1&ans2;
    }
    push_up(rt);
    return ret;
}
void work() {
    int n;
    scanf("%d", &n);
    int tot=0;
    for (int i = 0; i < n; ++i) {
        scanf("%d%d", &l[i], &r[i]);
        a[++tot] = l[i], a[++tot] = r[i];
    }
    sort(a+1,a+tot+1);
    tot = unique(a+1,a+tot+1)-a-1;

    b[1] = a[1]; int cnt = 1;
    for (int i = 2; i <= tot; ++i) {
        if (a[i]-a[i-1] == 1) b[++cnt] = a[i];
        else b[++cnt] = a[i-1] + 1, b[++cnt] = a[i];
    }

    build(1,1,cnt);
    int ans=0;
    for (int i = n-1; i >= 0; --i) {
        int pl = lower_bound(b+1, b+1+cnt, l[i]) - b,
            pr = lower_bound(b+1, b+1+cnt, r[i]) - b;
        if (!insert(1, pl, pr)) ++ans;
    }
    printf("%d\n", ans);
}
int main() {
    int T;
    scanf("%d", &T);
    while (T--) work();
    return 0;
}

Ver. 3

#include <stdio.h>
#include <algorithm>
#include <map>
#define maxn 100010
using namespace std;
typedef long long LL;
int a[maxn], l[maxn], r[maxn], fa[maxn], b[maxn];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
map<int,int> mp;
void work() {
    int n;
    scanf("%d", &n);
    int tot=0;
    for (int i = 0; i < n; ++i) {
        scanf("%d%d", &l[i], &r[i]);
        a[++tot] = l[i], a[++tot] = r[i];
    }
    sort(a+1,a+tot+1);
    tot = unique(a+1,a+tot+1)-a-1;

    b[1] = a[1]; int cnt = 1;
    for (int i = 2; i <= tot; ++i) {
        if (a[i]-a[i-1] == 1) b[++cnt] = a[i];
        else b[++cnt] = a[i-1] + 1, b[++cnt] = a[i];
    }
    for (int i=0; i<=cnt; ++i) fa[i] = i;

    int ans=0;
    for (int i = n-1; i >= 0; --i) {
        int pl = lower_bound(b+1, b+1+cnt, l[i]) - b,
            pr = lower_bound(b+1, b+1+cnt, r[i]) - b;
        if (find(pr)<pl) continue;
        ++ans;
        for (int j = find(pr); j >= pl; j = find(j-1)) {
            fa[find(j)] = find(pl-1);
        }
    }
    printf("%d\n", ans);
}
int main() {
    int T;
    scanf("%d", &T);
    while (T--) work();
    return 0;
}