2 2 1 10 4 9 3 5 1 2 3 4 5 2 4 6 8 10 3 6 9 12 15
1 1
开始用深搜，结果超时，换成O（n^3）的DP竟然过了，看来时间复杂度分析还是不过关唉。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
bool arr[21][502];
int dp[21][502];
int main(){
int n, k, t, minlen;
while(scanf("%d%d", &n, &k) == 2){
memset(arr, 0, sizeof(arr));
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= k; ++j){
scanf("%d", &t);
arr[i][t] = 1;
}
for(int i = 2; i <= n; ++i){
for(int j = 1; j <= 501; ++j){
if(i == 2 && arr[i][j]){
for(int k = 0; k < 501; ++k){
if(arr[i-1][k]){
if(dp[i][j] == 0) dp[i][j] = dp[i-1][k] + abs(k - j);
else if(dp[i-1][k] + abs(k - j) < dp[i][j])
dp[i][j] = dp[i-1][k] + abs(k - j);
}
}
continue;
}
if(arr[i][j]){
for(int k = 0; k < 501; ++k){
if(dp[i-1][k]){
if(dp[i][j] == 0) dp[i][j] = dp[i-1][k] + abs(k - j);
else if(dp[i-1][k] + abs(k - j) < dp[i][j])
dp[i][j] = dp[i-1][k] + abs(k - j);
}
}
}
}
}
minlen = 0;
for(int i = 1; i < 501; ++i)
if(dp[n][i]){
if(!minlen) minlen = dp[n][i];
else if(dp[n][i] < minlen) minlen = dp[n][i];
}
printf("%d\n", minlen);
}
return 0;
}
HDOJ4540 威威猫系列故事——打地鼠 【DP】,布布扣,bubuko.com
原文：http://blog.csdn.net/chang_mu/article/details/25397263