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package class42 ;
public class Problem_0265_PaintHouseII {
// costs[i][k] i号房子用k颜色刷的花费
// 要让0...N-1的房子相邻不同色
// 返回最小花费
public static int minCostII ( int [ ] [ ] costs ) {
int N = costs . length ;
if ( N = = 0 ) {
return 0 ;
}
int K = costs [ 0 ] . length ;
// 之前取得的最小代价、取得最小代价时的颜色
int preMin1 = 0 ;
int preEnd1 = - 1 ;
// 之前取得的次小代价、取得次小代价时的颜色
int preMin2 = 0 ;
int preEnd2 = - 1 ;
for ( int i = 0 ; i < N ; i + + ) { // i房子
int curMin1 = Integer . MAX_VALUE ;
int curEnd1 = - 1 ;
int curMin2 = Integer . MAX_VALUE ;
int curEnd2 = - 1 ;
for ( int j = 0 ; j < K ; j + + ) { // j颜色!
if ( j ! = preEnd1 ) {
if ( preMin1 + costs [ i ] [ j ] < curMin1 ) {
curMin2 = curMin1 ;
curEnd2 = curEnd1 ;
curMin1 = preMin1 + costs [ i ] [ j ] ;
curEnd1 = j ;
} else if ( preMin1 + costs [ i ] [ j ] < curMin2 ) {
curMin2 = preMin1 + costs [ i ] [ j ] ;
curEnd2 = j ;
}
} else if ( j ! = preEnd2 ) {
if ( preMin2 + costs [ i ] [ j ] < curMin1 ) {
curMin2 = curMin1 ;
curEnd2 = curEnd1 ;
curMin1 = preMin2 + costs [ i ] [ j ] ;
curEnd1 = j ;
} else if ( preMin2 + costs [ i ] [ j ] < curMin2 ) {
curMin2 = preMin2 + costs [ i ] [ j ] ;
curEnd2 = j ;
}
}
}
preMin1 = curMin1 ;
preEnd1 = curEnd1 ;
preMin2 = curMin2 ;
preEnd2 = curEnd2 ;
}
return preMin1 ;
}
}