解题思路

贪心

每个长条都不能落下，都需要被剪到，那么我们处理到全部长条的左右端点，按照右端点第一关键字，左端点第二关键字升序排序后，对于每块布条最好就是在终点前剪下，后面左端点比之前落下位置大的就要重新开刀。说白了就是看电源问题，挺典型的贪心模型。

#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }    while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();    return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 32000 + 10;
struct Node {
    int l, r;
    bool operator < (const Node& b) const {
        if (r != b.r)    return r < b.r;
        return l < b.l;
    }
}a[N];

int main() {
    int n = read();
    for (int i = 1; i <= n; ++i) {
        int st = read(), len = read();
        a[i].l = st, a[i].r = st + len;
    }
    sort(a + 1, a + 1 + n);
    int ans = 1, ed = a[1].r;
    for (int i = 2; i <= n; ++i)
        if (a[i].l >= ed)    ++ans, ed = a[i].r;
    write(ans), putchar(10);
    return 0;
}