題意:
(from luckycat)
解法:
參考 http://www.cppblog.com/rakerichard/archive/2010/02/19/108081.html。
TAG: DP,
注意:
程式碼:
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/** | |
* Tittle: 10271 - Chopsticks | |
* Author: Cheng-Shih, Wong | |
* Date: 2015/06/02 | |
*/ | |
// include files | |
#include <bits/stdc++.h> | |
using namespace std; | |
// definitions | |
#define FOR(i,a,b) for( int i=(a),_n=(b); i<=_n; ++i ) | |
#define clr(x,v) memset( x, v, sizeof(x) ) | |
#define pre(x) ((x-1)&1) | |
#define cur(x) ((x)&1) | |
#define INF 1e9 | |
#define N 5005 | |
#define K 1015 | |
// declarations | |
int t; | |
int k, n; | |
int len[N]; | |
// functions | |
int dp[2][N]; | |
int w( int i, int j ) | |
{ | |
int d = len[i]-len[j]; | |
return d*d; | |
} | |
void solve( void ) | |
{ | |
int ans; | |
clr( dp[0], 0 ); | |
FOR( i, 1, k ) { | |
FOR( j, 1, n ) { | |
if( i*3 > j ) dp[cur(i)][j] = INF; | |
else dp[cur(i)][j] = min( | |
dp[cur(i)][j-1], | |
dp[pre(i)][j-2]+w(j,j-1) | |
); | |
} | |
} | |
ans = INF; | |
FOR( i, k*3, n ) ans = min( ans, dp[cur(k)][i] ); | |
printf( "%d\n", ans ); | |
} | |
// main function | |
int main( void ) | |
{ | |
// input | |
scanf( "%d", &t ); | |
while( t-- ) { | |
scanf( "%d%d", &k, &n ); | |
k += 8; | |
for( int i=n; i>=1; --i ) | |
scanf( "%d", &len[i] ); | |
solve(); | |
} | |
return 0; | |
} |
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