POJ 1260 Pearls （斜率DP）题解

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\n",dp[n]);
	}
    return 0;
}