hdu 4824 Disk Schedule(双调欧几里得旅行商问题)

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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;}

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