POJ1180 Batch Scheduling -斜率优化DP

题解

　　将费用提前计算可以得到状态转移方程： $F_i = \min(F_j + sumT_i * (sumC_i - sumC_j) + S \times (sumC_N - sumC_j)$

　　把方程进行分离， 得到 $S\times sumC_j  + F_j = sumT_i \times sumC_j + F_i - S \times sumC_N$。

　　 将等号左边看成纵坐标， $sumC_j$看成横坐标， $sumT_i$为斜率来进行斜率优化。

　　由于 $sumT_i$是递增的， 即斜率是递增的， 维护一个单调队列， 第一个斜率大于$sumT_i$的端点就为决策点

　　斜率优化dp还是很套路的

代码

　　

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #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;

 const int N = 1e4 + 1e3;

 int n, m, sumT[N], sumC[N], S, f[N], q[N];

 int cross(int a, int b, int c) { //点积
     int ax = sumC[a], bx = sumC[b], cx = sumC[c];
     int ay = f[a], by = f[b], cy = f[c];
     int x = bx - ax, xx = cx - ax, y = by - ay, yy = cy - ay;
     return x * yy - xx * y;
 }// 向量ab, ac

 double calk(int a, int b) {
     int ax = sumC[a], bx = sumC[b], ay = f[a], by = f[b];
     return 1.0 * (by - ay) / (bx - ax);
 }

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

 int main()
 {
     n = rd; S = rd;
     rep(i, , n) {
         int t = rd, c = rd;
         sumT[i] = sumT[i - ] + t;
         sumC[i] = sumC[i - ] + c;
     }
     memset(f, 0x3f, sizeof(f));
     int l = , r = ;
     q[] = f[] = ;
     rep(i, , n) {
         while(l < r && calk(q[l], q[l + ]) <= S + sumT[i]) l++;
         if(l <= r) f[i] = f[q[l]] + sumT[i] * (sumC[i] - sumC[q[l]]) + S * (sumC[n] - sumC[q[l]]);
         while(l < r && cross(q[r - ], q[r], i) <= ) r--;
         q[++r] = i;
     }
     printf("%d\n",f[n]);
 }