题目链接:hdu 4824 Disk Schedule

题目大意：中文题。

解题思路：需要的时，很明显每到一层是要读取一次数据的，但是因为需要返回00,所以有些层的数据可以在返回的过程中读取会较优。于是转化成了双调欧几里得旅行商问题。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;
const int N = 1005;
const int INF = 0x3f3f3f3f;
int n, p, d[N], dp[N][N];

int dis (int a, int b) {
    int tmp = abs(d[a] - d[b]);
    return min(tmp, 360 - tmp);
}

void init () {
    int a;

    scanf("%d", &n);
    d[1] = 0;
    for (int i = 2; i <= n + 1; i++) {
        scanf("%d%d", &a, &d[i]);

        if (i == n + 1)
            p = a;
    }
    dp[2][1] = dis(1, 2);
}

int solve () {

    for (int i = 3; i <= n + 1; i++) {  
        dp[i][i-1] = INF;  

        for (int j = 1; j < i-1; j++) {  
            dp[i][i-1] = min(dp[i][i-1], dp[i-1][j] + dis(i, j));  
            dp[i][j] = dp[i-1][j] + dis(i, i-1);  
        }  
    }  

    int ans = INF;  
    for (int i = 1; i <= n; i++)  
        ans = min(ans, dp[n+1][i] + dis(n+1, i));  
    return ans;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int i = 1; i <= cas; i++) {
        init();
        printf("%d\n", solve() + p * 800 + 10 * n);
    }
    return 0;
}