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题意:n个饭店在一条直线上,给了它们的坐标,现在要建造m个停车场,饭店没有停车场的要到最近的停车场,问所有饭店到停车场的最短距离
分析:易得区间(i, j)的最短距离和一定是建在(i + j) / 2的饭店,预处理出(i, j)的距离和sum[i][j],mark[i][j] 表示区间的最优停车场的位置,mid[i][j]表示(i + j) / 2。状态转移方程:dp[i][j] = max (dp[k-1][j-1] + sum[k][i]);
收获:学习递归打印路径
代码:
/************************************************ * Author :Running_Time * Created Time :2015-8-29 16:42:33 * File Name :UVA_662.cpp ************************************************/ #include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std; #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 2e2 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; int dp[N][33]; int sum[N][N], d[N], mark[N][N], mid[N][N]; int n, m; int idx; void print(int i, int j) { if (i < 1 || j < 1) return ; print (mark[i][j]-1, j-1); printf ("Depot %d at restaurant %d serves restaurant", ++idx, mid[mark[i][j]][i]); if (mark[i][j] == i) { printf (" %d\n", i); return ; } else printf ("s %d to %d\n", mark[i][j], i); } int main(void) { int cas = 0; while (scanf ("%d%d", &n, &m) == 2) { if (!n && !m) break; for (int i=1; i<=n; ++i) scanf ("%d", &d[i]); memset (sum, 0, sizeof (sum)); for (int i=1; i<=n; ++i) { mid[i][i]= i; for (int j=i+1; j<=n; ++j) { int mm = (i + j) >> 1; mid[i][j] = mm; for (int k=i; k<=j; ++k) { sum[i][j] += abs (d[mm] - d[k]); } } } memset (dp, INF, sizeof (dp)); memset (dp[0], 0, sizeof (dp[0])); for (int i=1; i<=n; ++i) { for (int j=1; j<=m; ++j) { for (int k=1; k<=i; ++k) { int tmp = dp[k-1][j-1] + sum[k][i]; if (dp[i][j] >= tmp) { dp[i][j] = tmp; mark[i][j] = k; } } } } printf ("Chain %d\n", ++cas); idx = 0; print (n, m); printf ("Total distance sum = %d\n\n", dp[n][m]); } return 0; }
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原文地址:http://www.cnblogs.com/Running-Time/p/4773925.html