普及练习场 线性动态规划 石子合并

题目链接

题意理解

首先这道题目不能用这样的贪心来做：对于环中的和最小的两个石子堆，先合并。因为这样做可能会导致在有的时候，不止一个和最小的两个石子堆，你在这里合并了之后，会影响后继的合并状态。这与之前可以用贪心写的不一样，是因为之前是可以任意合并两堆石子，而这里是环形中可以合并相邻的两堆石子。然后需要注意一下断环成链的技巧。注意点大部分都写到注释里面了。

代码

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
static int N;
static final int maxn = 220;
static int[] a = new int[maxn];
static int[] sum = new int[maxn];
static int[][] dp1 = new int[maxn][maxn];
static int[][] dp2 = new int[maxn][maxn];
public static void main(String[] args) {
FastScanner fs = new FastScanner();
N = fs.nextInt();
for(int i = 1; i <= N; i++) {
a[i] = fs.nextInt();
a[i+N] = a[i];
sum[i] = sum[i-1] + a[i];
}
for(int i = N + 1; i <= 2 * N; i++) {
sum[i] = sum[i-1] + a[i];
}

// dp[l][r] = min{dp[l][k] + dp[k+1][r] + sum(l, r)}
// 此时正确的枚举方式应该是：先枚举一遍r-l=1的，再枚举一遍r-l=2的,……

for(int len = 1; len <= N; len++) {
for(int l = 1; l <= 2 * N; l++) {
int r = l + len;
if(r > N * 2) {
break;
}
int min = 0x3f3f3f3f;
int max = -1;
for(int k = l; k < r; k++) {
min = Math.min(min, dp1[l][k] + dp1[k+1][r] + sum(l, r));
max = Math.max(max, dp2[l][k] + dp2[k+1][r] + sum(l, r));
}
dp1[l][r] = min;
dp2[l][r] = max;
}
}
int minScore = 0x3f3f3f3f;
int maxScore = -1;
for(int i = 1; i <= N; i++) {
minScore = Math.min(minScore, dp1[i][i+N-1]);
maxScore = Math.max(maxScore, dp2[i][i+N-1]);
}
System.out.println(minScore);
System.out.println(maxScore);
}
public static int sum(int l, int r) {
return sum[r] - sum[l-1];
}

public static class FastScanner {
private BufferedReader br;
private StringTokenizer st;

public FastScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}

public String nextToken() {
while(st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
// TODO: handle exception
}
}
return st.nextToken();
}

public int nextInt() {
return Integer.valueOf(nextToken());
}
}
}