Codevs 1643 & 3027 线段覆盖

1643

题意

给定若干条线段，问最多可以安排多少条使得没有重合。

思路

贪心，同安排schedule，按结束时间早的排序。

Code

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define maxn 1000010
using namespace std;
typedef long long LL;
int n;
struct Seg {
    int l, r;
    bool operator < (const Seg& s) const { return r < s.r; }
}seg[maxn];
void work() {
    for (int i = 0; i < n; ++i) {
        scanf("%d%d", &seg[i].l, &seg[i].r);
        if (seg[i].l > seg[i].r) swap(seg[i].l, seg[i].r);
    }
    sort(seg, seg+n);
    int r = -inf, cnt = 0;
    for (int i = 0; i < n; ++i) {
        if (seg[i].l >= r) ++cnt, r = seg[i].r;
    }
    printf("%d\n", cnt);
}
int main() {
    scanf("%d", &n); work();
    return 0;
}

3027

题意

给定若干条线段，每条线段都有各自的价值，问怎样安排使得不重叠且总价值最大。

思路

dp

Code

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define maxn 1010
using namespace std;
typedef long long LL;
struct Seg {
    int l, r; LL w;
    bool operator < (const Seg& s) const { return r < s.r; }
}seg[maxn];
LL dp[maxn];
int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) scanf("%d%d%d", &seg[i].l, &seg[i].r, &seg[i].w);
    sort(seg, seg+n);
    dp[0] = seg[0].w;
    LL ans = dp[0];
    for (int i = 1; i < n; ++i) {
        dp[i] = 0;
        for (int j = 0; j < i; ++j) {
            if (seg[j].r <= seg[i].l) dp[i] = max(dp[i], dp[j]);
        }
        dp[i] += seg[i].w;
        ans = max(ans, dp[i]);
    }
    printf("%lld\n", ans);
    return 0;
}