标签：second why iso into cat ant first contain operation
Another OCD Patient
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2259 Accepted Submission(s):
795
Problem Description
Xiaoji is an OCD (obsessive-compulsive disorder)
patient. This morning, his children played with plasticene. They broke the
plasticene into N pieces, and put them in a line. Each piece has a volume Vi.
Since Xiaoji is an OCD patient, he can‘t stand with the disorder of the volume
of the N pieces of plasticene. Now he wants to merge some successive pieces so
that the volume in line is symmetrical! For example, (10, 20, 20, 10), (4,1,4)
and (2) are symmetrical but (3,1,2), (3, 1, 1) and (1, 2, 1, 2) are
not.
However, because Xiaoji‘s OCD is more and more serious, now he has a
strange opinion that merging i successive pieces into one will cost ai. And he
wants to achieve his goal with minimum cost. Can you help him?
By the
way, if one piece is merged by Xiaoji, he would not use it to merge again. Don‘t
ask why. You should know Xiaoji has an OCD.
Input
The input contains multiple test cases.
The
first line of each case is an integer N (0 < N <= 5000), indicating the
number of pieces in a line. The second line contains N integers Vi, volume of
each piece (0 < Vi <=10^9). The third line contains N integers ai (0
< ai <=10000), and a1 is always 0.
The input is terminated by
N = 0.
Output
Output one line containing the minimum cost of all
operations Xiaoji needs.
Sample Input
5
6 2 8 7 1
0 5 2 10 20
0
Sample Output
10
题意：一个长为n的序列，可对任意长度的序列进行合并，代价为ai，要求花最小代价将其构造成回文序列
对于回文序列，可以找到两个点i，j（i < j）使1-i的和与j-n的和相等，那么就可以预处理出这样的点的对数，存在 i‘ < i < j < j‘ ，L[i] = R[j]，而L[i‘] = R[j‘]，所以有L[i]-L[i‘] = R[j]-R[j‘]，于是序列被分为5段区间
可以从中间向两边进行dp，转移方程为 dp[i] = min(dp[i]，dp[j]+p[len(i‘，i)]+p[len(j，j‘)])
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#define ll long long
using namespace std;
const int N = 5005;
int t, n, m, k, ans;
ll v[N], p[N], lt[N], rt[N];
ll L[N], R[N], dp[N];
int main(){
while(~scanf("%d", &n) && n){
L[0] = R[n+1] = 0;
for(int i = 1; i <= n; i ++){
scanf("%I64d", &v[i]);
L[i] = L[i-1]+v[i];
}
for(int i = 1; i <= n; i ++){
scanf("%I64d", &p[i]);
}
for(int i = n; i; i --){
R[i] = R[i+1]+v[i];
}
ans = 0;
int l = 1, r = n;
while(l <= r+1){ //WA点，注意最中间一段区域的处理
if(L[l-1] == R[r+1]){
lt[++ans] = l;
rt[ans] = r;
l ++, r --;
}
while(l <= r+1 && L[l-1] < R[r+1])
l ++;
while(l <= r+1 && L[l-1] > R[r+1])
r --;
}
for(int i = ans; i; i --){
dp[i] = p[rt[i]-lt[i]+1];
for(int j = i+1; j <= ans; j ++){
dp[i] = min(dp[i], dp[j]+p[lt[j]-lt[i]]+p[rt[i]-rt[j]]);
}
}
printf("%I64d\n", dp[1]);
}
return 0;
}
View Code
hdu 4960（预处理+DP）
标签：second why iso into cat ant first contain operation
原文地址：https://www.cnblogs.com/microcodes/p/12877892.html
