玲珑杯 1066（“玲珑杯”ACM比赛 Round #6）（区间DP+四边形不等式优化）

1066 - Buy Candy

Time Limit：1s Memory Limit：1024MByte

Submissions：185Solved：39

DESCRIPTION

PigVan ( Mr.Van's pet ) encounter a hard problem in his school. There are n groups children sit in a line. However, all groups dislike each other. Each group have a value which means how others dislike him. And PigVan wants to make all of them become one group. He can only merge two adjacent group into one. And if he wants to merge two groups of children, he needs to buy f(s) candies to them. ( s is the sum of their dislike value andf is a polynomial function of s ).

He wonder the minimal cost he will spend on candies.

INPUT

The first line is a single integer T, indicating the number of test cases.
For each test case:
In the first line, there are an integer $n(1\le n\le 1000)$
In the second line, there are $n$ integers. The $i$$th$ integer $s$$i$ (1 ≤ $s$$i$ ≤ 40), indicating the dislike value of the $i$$th$ child.
In the third line, there are an integer $m(1\le m\le 4)$In the next line, there are $m+1$ integers a0 , a1 ,..., am.
The polynomial function f(x) = a0 + a1x + a2x2 +...+ amxm$(1\le$ai$\le 5)$

OUTPUT

For each test case, output an integer indicating the answer.

SAMPLE INPUT

153 1 8 9 922 1 2

SAMPLE OUTPUT

2840

SOLUTION

“玲珑杯”ACM比赛 Round #6

#include <stdio.h>#include <string.h>#include <vector>#define LL long longusing namespace std;const int maxn = 1005;const long long inf = 1LL << 62;LL dp[maxn][maxn];int p[maxn][maxn];LL a[1005] = { 0 };LL b[55];int n, m;LL jici(LL x, int y){LL ret = 1;for (int i = 1; i <= y; i++)ret *= x;return ret;}LL fun(LL x){LL ret = 0;for (int i = 0; i <= m; i++)ret += jici(x, i)*b[i];return ret;}int main(){int T;scanf("%d", &T);while (T--){scanf("%d", &n);for (int i = 1; i <= n; i++){scanf("%lld", &a[i]);a[i] += a[i - 1];dp[i][i] = 0;p[i][i] = i;}scanf("%d", &m);for (int i = 0; i <= m; i++)scanf("%lld", &b[i]);LL sum = 0;for (int len = 1; len <= n - 1; len++){for (int i = 1; i + len <= n; i++){int j = i + len;LL F = fun(a[j] - a[i-1]);dp[i][j] = inf;for (int k = p[i][j - 1]; k <= p[i + 1][j]; k++){if (dp[i][j] > dp[i][k] + dp[k + 1][j]+F){dp[i][j] = dp[i][k] + dp[k + 1][j] + F;p[i][j] = k;}}}}printf("%lld\n", dp[1][n]);}return 0;}