清理班次

农民约翰正在指挥他的N头牛进行清理工作。

他将一天划分为了T个班次（1~T）。

每头牛都只能在一天中的某一个时间段内进行不间断的工作。

你需要帮助约翰排列出一个合理的奶牛的清理班次，使得每个班次都有奶牛在进行清理，而且动用的奶牛数量可以尽可能的少。

输入格式
第1行：两个空格隔开的整数N和T。

第2…N+1行：第i+1行包含两个整数，分别表示第i头牛可以进行工作的开始时间和结束时间。

输出格式
输出一个整数，表示在每个班次都有奶牛清理的情况下，所需的奶牛最小数量。

如果无法做到每个班次都有奶牛清理，则输出-1。

数据范围
1≤N≤25000,
1≤T≤10^6

输入样例：
3 10
1 7
3 6
6 10
输出样例：
2

方法1,：贪心
数据范围小，不需要离散化

#include<bits/stdc++.h>

using namespace std;
const int N = 1e6 + 10;
int Next[N], n, t, l, r, p = 1, ans;

int main() {
    scanf("%d%d", &n, &t);
    for (int i = 1; i <= n; i++) {
        scanf("%d%d", &l, &r);
        Next[l] = max(Next[l], r);
    }
    for (int i = 2; i <= t; i++)
        Next[i] = max(Next[i], Next[i - 1]);
    while (p <= t && Next[p] >= p)p = Next[p] + 1, ans++;
    p > t ? printf("%d\n", ans) : puts("-1");
    return 0;
}

(离散反而更慢)

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#include<bits/stdc++.h>

#define pii pair<int,int>
#define fi first
#define se second
using namespace std;
const int N = 1e5 + 10;
int Next[N], n, t, l, r, p = 1, ans;
vector<int> seq;
vector<pii > cow;
unordered_map<int, int> num;

int main() {
    scanf("%d%d", &n, &t);
    seq.push_back(1), seq.push_back(t);
    for (int i = 1; i <= n; i++) {
        scanf("%d%d", &l, &r);
        cow.emplace_back(l, r);
        seq.push_back(l), seq.push_back(r), seq.push_back(r + 1);
    }
    sort(seq.begin(), seq.end());
    seq.erase(unique(seq.begin(), seq.end()), seq.end());
    for (int i = 0; i < seq.size(); i++)num[seq[i]] = i + 1;
    for (int i = 0; i < n; i++)
        Next[num[cow[i].fi]] = max(Next[num[cow[i].fi]], num[cow[i].se]);
    for (int i = 2; i <= seq.size(); i++)
        Next[i] = max(Next[i], Next[i - 1]);
    while (p <= num[t] && Next[p] >= p)p = Next[p] + 1, ans++;
    p > num[t] ? printf("%d\n", ans) : puts("-1");
    return 0;
}

方法2：dp+线段树优化
不离散

#include<bits/stdc++.h>

using namespace std;
const int N = 25e3 + 10, M = 1e6 + 10, INF = 0x3f3f3f3f;

struct T {
    int l, r, dat;

    bool operator<(const T w) const {
        return r < w.r;
    }
} a[N], t[M << 3];

int f[M];

void build(int p, int l, int r) {
    t[p].l = l, t[p].r = r;
    if (l == r) {
        t[p].dat = INF;
        return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    t[p].dat = min(t[p << 1].dat, t[p << 1 | 1].dat);
}

void change(int p, int x, int v) {
    if (t[p].l == t[p].r) {
        t[p].dat = v;
        return;
    }
    int mid = (t[p].l + t[p].r) >> 1;
    change(x <= mid ? p << 1 : p << 1 | 1, x, v);
    t[p].dat = min(t[p << 1].dat, t[p << 1 | 1].dat);
}

int ask(int p, int l, int r) {
    if (l <= t[p].l && r >= t[p].r)return t[p].dat;
    int mid = (t[p].l + t[p].r) >> 1;
    int val = INF;
    if (l <= mid)val = min(val, ask(p << 1, l, r));
    if (r > mid)val = min(val, ask(p << 1 | 1, l, r));
    return val;
}

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d%d", &a[i].l, &a[i].r);
    sort(a + 1, a + 1 + n);
    build(1, 0, m);
    change(1, 0, 0);
    memset(f, 0x3f, sizeof(f));
    f[0] = 0;
    for (int i = 1; i <= n; i++) {
        int l = a[i].l, r = a[i].r;
        int val = ask(1, l - 1, r - 1) + 1;
        if (val < f[r]) {
            change(1, r, val);
            f[r] = val;
        }
    }
    f[m] == INF ? puts("-1") : printf("%d\n", f[m]);
    return 0;
}

离散化后快一些

#include<bits/stdc++.h>

using namespace std;
const int N = 25e3 + 10, INF = 0x3f3f3f3f;

struct T {
    int l, r, dat;

    bool operator<(const T w) const {
        return r < w.r;
    }
} a[N], t[N * 12];

unordered_map<int, int> num;
vector<int> seq;
int f[N * 3];

void build(int p, int l, int r) {
    t[p].l = l, t[p].r = r;
    if (l == r) {
        t[p].dat = INF;
        return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    t[p].dat = min(t[p << 1].dat, t[p << 1 | 1].dat);
}

void change(int p, int x, int v) {
    if (t[p].l == t[p].r) {
        t[p].dat = v;
        return;
    }
    int mid = (t[p].l + t[p].r) >> 1;
    change(x <= mid ? p << 1 : p << 1 | 1, x, v);
    t[p].dat = min(t[p << 1].dat, t[p << 1 | 1].dat);
}

int ask(int p, int l, int r) {
    if (l <= t[p].l && r >= t[p].r)return t[p].dat;
    int mid = (t[p].l + t[p].r) >> 1;
    int val = INF;
    if (l <= mid)val = min(val, ask(p << 1, l, r));
    if (r > mid)val = min(val, ask(p << 1 | 1, l, r));
    return val;
}

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d%d", &a[i].l, &a[i].r);
        seq.push_back(a[i].l - 1);
        seq.push_back(a[i].l);
        seq.push_back(a[i].r);
    }
    seq.push_back(0), seq.push_back(m);
    sort(a + 1, a + 1 + n);
    sort(seq.begin(), seq.end());
    seq.erase(unique(seq.begin(), seq.end()), seq.end());
    for (int i = 0; i < seq.size(); i++)num[seq[i]] = i;
    build(1, 0, seq.size());
    change(1, 0, 0);
    memset(f, 0x3f, sizeof(f));
    f[0] = 0;
    for (int i = 1; i <= n; i++) {
        int l = num[a[i].l], r = num[a[i].r];
        int val = ask(1, l - 1, r - 1) + 1;
        if (val < f[r])change(1, r, val), f[r] = val;
    }
    f[num[m]] == INF ? puts("-1") : printf("%d\n", f[num[m]]);
    return 0;
}