标签：dp
5 6 2 8 7 1 0 5 2 10 20 0
10HintIn the sample, there is two ways to achieve Xiaoji‘s goal. [6 2 8 7 1] -> [8 8 7 1] -> [8 8 8] will cost 5 + 5 = 10. [6 2 8 7 1] -> [24] will cost 20.
分析：由于题目要求把原数列变成回文数列，所以只需要找出原数组中的关键点，然后利用这些关键点进行dp。
所谓的关键点就是指数对（i，j）满足v1+v2+……+vi = vj+……+vn。因为关键点i和j是一一对应的，所以一维dp就能解决问题。则dp[i] = min(dp[j-1]+a[l_num[j,i]]+a[r_num[j,i]]|1<=j<=i),其中l_num[j,i]表示左边第j个关键点到第i个关键点的元素个数的总数，r_num[j,i]表示右边第j个关键点到第i个关键点的元素个数的总数
#include<cstdio>
#include<algorithm>
using namespace std;
#define INF 0x3fffffff
const int N = 5005;
int n, v[N], a[N];
int dp[N];
int L[N], R[N], num; //L[]和R[]记录关键点
void Init()
{
num = 0;
int l = 1, r = n, l_num = 1, r_num = 1;
int l_val = v[1], r_val = v[n];
while(l < r) {
if(l_val < r_val) {
l_val += v[++l];
l_num++;
}
else if(l_val > r_val) {
r_val += v[--r];
r_num++;
}
else {
L[++num] = l_num;
R[num] = r_num;
l_num = r_num = 1;
l_val = v[++l];
r_val = v[--r];
}
}
}
int main()
{
while(~scanf("%d",&n) && n) {
for(int i = 1; i <= n; ++i) scanf("%d",&v[i]);
for(int i = 1; i <= n; ++i) scanf("%d",&a[i]);
Init();
for(int i = 1; i <= num; i++) {
int num1 = 0, num2 = 0;
dp[i] = INF;
for(int j = i; j >= 1; --j) {
num1 += L[j];
num2 += R[j];
dp[i] = min(dp[i], dp[j-1] + a[num1] + a[num2]);
}
}
int ans = a[n], cnt = n;
for(int i = 1; i <= num; ++i) {
cnt -= L[i] + R[i];
ans = min(ans, dp[i] + a[cnt]);
}
printf("%d\n", ans);
}
return 0;
}hdu 4960 Another OCD Patient（dp）2014多校训练第9场,布布扣,bubuko.com
hdu 4960 Another OCD Patient（dp）2014多校训练第9场
标签：dp
原文地址：http://blog.csdn.net/lyhvoyage/article/details/38708245
