BZOJ 2726 [SDOI2012] 任务安排

题解

转移方程与我的上一篇题解一样 ： $S imes sumC_j  + F_j = sumT_i imes sumC_j + F_i - S imes sumC_N$。

分离成：$S imes sumC_j  + F_j = sumT_i imes sumC_j + F_i - S imes sumC_N$

不同的是， 时间可能为 负数（出题人解释：不要把时间看的这么狭义。

所以$sumT_i$不是递增。

所以我们不能在队首弹出斜率比 $sumT_i$小的数， 只能用一个数据结构来维护并查询, 我当然是用了好玩的set

但是队尾还是需要维护下凸壳。

有一个坑点：用叉积来弹出队尾可能爆longlong， 开double可以过QAQ

代码

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#define rd read()
#define rep(i,a,b) for( int i = (a); i <= (b); ++i)
#define per(i,a,b) for( int i = (a); i >= (b); --i)
using namespace std;
#define ll long long
typedef pair<double, ll>P;

const ll N = 1e6;
const ll inf = 1e18;
const double eps = 1e-6;

ll S, sumT[N], sumC[N], q[N], n;
ll f[N];

set<P>st;

ll read() {
    ll X = 0, p = 1; char c = getchar();
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    return X * p;
}

ll cross(ll a, ll b, ll c) { //叉积
    ll ax = sumC[a], bx = sumC[b], cx = sumC[c];
    ll ay = f[a], by = f[b], cy = f[c];
    ll x = bx - ax, y = by - ay, xx = cx - ax, yy = cy - ay;
    if(fabs(1LL * x * yy - 1LL * xx * y) < eps) return 0;
    return (long double)x * yy > (long double)xx * y;
}
    
double calc(ll a, ll b) {
    ll ax = sumC[a], bx = sumC[b];
    ll ay = f[a], by = f[b];
    if(bx - ax == 0) return by > ay ? inf : -inf;
    else return 1.0 * (by - ay) / (bx - ax);
}

int main()
{
    n = rd; S = rd;
    rep(i, 1, n) {
        ll t = rd, c = rd;
        sumT[i] = sumT[i - 1] + t;
        sumC[i] = sumC[i - 1] + c;
    }
    ll l = 1, r = 1;
    q[1] = f[0] = 0;
    set<P>::iterator it;
    st.insert(P(-inf, 0));
    rep(i, 1, n) {
        double k = S + sumT[i];
        P fd = P(k, 0);
        it = st.lower_bound(fd);//查找最小的斜率比sumT[i]大的
        it--;
        ll p = (*it).second;
        f[i] = f[p] + 1LL * sumT[i] * (sumC[i] - sumC[p]) + 1LL * S * (sumC[n] - sumC[p]);
        while(l < r && cross(q[r - 1], q[r], i) == 0) {
            st.erase(P(calc(q[r - 1], q[r]), q[r]));
            r--;
        }
        q[++r] = i;
        st.insert(P(calc(q[r - 1], q[r]), i));
    }
    printf("%lld
", f[n]);
    fclose(stdin); fclose(stdout);
}