思路：
直接DP也能做，这里用斜率DP。
dp[i] = min{ dp[j] + ( sum[i] - sum[j] + 10 )*pr[i]} ;
k<j<i => dp[j] - dp[k] <pr[i]*( sum[j] - sum[k] ）
再套模板
#include<queue>
#include<cstring>
#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<vector>
#include<cstdio>
#include<iostream>
#include<algorithm>
#define ll long long
const int N = 1000+5;
using namespace std;
int pr[N],sum[N],dp[N],q[N];
int up(int x,int y){
return dp[x] - dp[y];
}
int down(int x,int y){
return sum[x] - sum[y];
}
int main(){
int n,m,T;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(int i = 1;i <= n;i++) scanf("%d%d",&sum[i],&pr[i]);
for(int i = 2;i <= n;i++) sum[i] += sum[i-1];
int head,tail;
head = tail = 0;
dp[0] = 0;
q[tail++] = 0;
for(int i = 1;i <= n;i++){
while(head+1 < tail && up(q[head+1],q[head]) <= pr[i]*down(q[head+1],q[head])){
head++;
}
dp[i] = dp[q[head]] + (sum[i] - sum[q[head]] + 10)*pr[i];
while(head + 1 < tail && up(i,q[tail - 1])*down(q[tail - 1],q[tail - 2]) <= up(q[tail - 1],q[tail - 2])*down(i,q[tail - 1])){
tail--;
}
q[tail++] = i;
}
printf("%lld
",dp[n]);
}
return 0;
}