BZOJ 2726 [SDOI2012] 任务安排 - 斜率优化dp

题解

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

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

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-;

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

 set<P>st;

 ll read() {
     ll X = , p = ; char c = getchar();
     for(; c > '' || c < ''; c = getchar()) if(c == '-') p = -;
     for(; c >= '' && c <= ''; c = getchar()) X = X *  + c - '';
     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 ;
     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 == ) return by > ay ? inf : -inf;
     else return 1.0 * (by - ay) / (bx - ax);
 }

 int main()
 {
     n = rd; S = rd;
     rep(i, , n) {
         ll t = rd, c = rd;
         sumT[i] = sumT[i - ] + t;
         sumC[i] = sumC[i - ] + c;
     }
     ll l = , r = ;
     q[] = f[] = ;
     set<P>::iterator it;
     st.insert(P(-inf, ));
     rep(i, , n) {
         double k = S + sumT[i];
         P fd = P(k, );
         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 - ], q[r], i) == ) {
             st.erase(P(calc(q[r - ], q[r]), q[r]));
             r--;
         }
         q[++r] = i;
         st.insert(P(calc(q[r - ], q[r]), i));
     }
     printf("%lld\n", f[n]);
     fclose(stdin); fclose(stdout);
 }