农夫约翰雇佣他的N头奶牛帮他进行牛棚的清理工作。
他将全天分为了很多个班次,其中第M个班次到第E个班次(包括这两个班次)之间必须都有牛进行清理。
这N头牛中,第 i 头牛可以从第ai个班次工作到第bi个班次,同时,它会索取ci的佣金。
请你安排一个合理的清理班次,使得[M,E]时间段内都有奶牛在清理,并且所需支付给奶牛的报酬最少。
输入格式
第1行:包含三个整数N,M和E。
第2…N+1行:第i+1行包含三个整数ai,bi,ci。
输出格式
输出一个整数,表示所需的最少佣金。
如果无法做到在要求时间段内都有奶牛清理,则输出-1。
数据范围
1≤N≤10000,
0≤M,E≤86399,
M≤ai≤bi≤E,
0≤ci≤500000
输入样例:
3 0 4
0 2 3
3 4 2
0 0 1
输出样例:
5
#include<bits/stdc++.h>
using namespace std;
const int N = 1e4 + 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, e;
scanf("%d%d%d", &n, &m, &e);
for (int i = 1; i <= n; i++) {
scanf("%d%d%d", &a[i].l, &a[i].r, &a[i].dat);
seq.push_back(a[i].l - 1);
seq.push_back(a[i].l);
seq.push_back(a[i].r);
}
seq.push_back(m - 1), seq.push_back(e);
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) + +a[i].dat;
if (val < f[r])change(1, r, val), f[r] = val;
}
f[num[e]] == INF ? puts("-1") : printf("%d\n", f[num[e]]);
return 0;
}
不离散更快
#include<bits/stdc++.h>
using namespace std;
const int N = 1e4 + 10, M = 86400, 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, e;
scanf("%d%d%d", &n, &m, &e);
for (int i = 1; i <= n; i++) {
scanf("%d%d%d", &a[i].l, &a[i].r, &a[i].dat);
a[i].l -= m - 1, a[i].r -= m - 1;
}
sort(a + 1, a + 1 + n);
build(1, 0, e - m + 1);
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) + a[i].dat;
if (val < f[r]) change(1, r, val), f[r] = val;
}
f[e - m + 1] == INF ? puts("-1") : printf("%d\n", f[e - m + 1]);
return 0;
}