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 = 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;
}

int main()
{
    n = rd; S = rd;
    rep(i, 1, n) {
        int t = rd, c = rd;
        sumT[i] = sumT[i - 1] + t;
        sumC[i] = sumC[i - 1] + c;
    }
    memset(f, 0x3f, sizeof(f));
    int l = 1, r = 1;
    q[1] = f[0] = 0;
    rep(i, 1, n) {
        while(l < r && calk(q[l], q[l + 1]) <= 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 - 1], q[r], i) <= 0) r--;
        q[++r] = i;
    }
    printf("%d\n",f[n]);
}