易错点：

• f[i]=min(f[j]+任务批(j~i)的总价格（sumT[i]*(sumC[i]-sumC[j])）+无限远点的弧光灯（启动费用*(sumT[n]-sumT[j])）.

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int MAXN=5200;
int f[MAXN],sumT[MAXN],sumC[MAXN];
int main(){
    memset(f,0x3f,sizeof(f));
    int n,s;
    scanf("%d%d",&n,&s);
    for(int i=1;i<=n;i++){
        int t,c;
        scanf("%d%d",&t,&c);
        sumT[i]=sumT[i-1]+t;
        sumC[i]=sumC[i-1]+c;
    }
    f[0]=0;
    for(int i=1;i<=n;i++){
        for(int j=0;j<i;j++){
            f[i]=min(f[i],f[j]
            +sumT[i]*(sumC[i]-sumC[j])
            +s*(sumC[n]-sumC[j])
            );
        }
    }
    printf("%d\n",f[n]);
    return 0;
}