uva 662(区间dp)

来源：互联网发布：宠物美容剪刀知乎编辑：程序博客网时间：2024/04/28 14:45

题意：有n个饭店，要在n个饭店中选k个建立补给站给距离最近的饭店补给，给出n个饭店的位置，问补给站建在哪k个饭店并且将饭店和自己的补给站的距离和输出。

题解：f[i][j]代表在前j个饭店建i个补给站的距离和。先用一个dis[i][j]表示从i到j建立1个补给站的总距离(一定是(i + j) / 2这个距离其他几个点距离和最小)，那么状态转移方程就是f[i][j] = min{f[i][j]，f[i - 1][k] + dis[k + 1][j]}。最后根据f数组递归打印路径。

#include <stdio.h>#include <string.h>#include <math.h>const int N = 205;const int INF = 0x3f3f3f3f;int n, k, p[N];int f[N][N], dis[N][N];void print_path(int r, int k) {if (k == 1) {if (r == 1)printf("Depot 1 at restaurant 1 serves restaurant 1\n");elseprintf("Depot 1 at restaurant %d serves restaurants 1 to %d\n", (1 + r) / 2, r);return;}for (int i = r - 1; i >= 1; i--) {if (f[k][r] == f[k - 1][i] + dis[i + 1][r]) {print_path(i, k - 1);if (i + 1 == r)printf("Depot %d at restaurant %d serves restaurant %d\n", k, r, r);elseprintf("Depot %d at restaurant %d serves restaurants %d to %d\n", k, (i + 1 + r) / 2, i + 1, r);break;}}}int main() {int cas = 1;while (scanf("%d%d", &n, &k) && (n + k)) {memset(f, INF, sizeof(f));for (int i = 1; i <= n; i++)scanf("%d", &p[i]);for (int i = 1; i <= n; i++) {for (int j = 1; j <= i; j++) {dis[j][i] = 0;int mid = p[(j + i) / 2];for (int k = j; k <= i; k++) {dis[j][i] += fabs(p[k] - mid);}}}for (int i = 1; i <= n; i++)f[1][i] = dis[1][i];for (int i = 2; i <= k; i++) {for (int j = i; j <= n; j++) {for (int k = i - 1; k <= j; k++) {f[i][j] = f[i][j] < f[i - 1][k] + dis[k + 1][j] ? f[i][j] : f[i - 1][k] + dis[k + 1][j];}}}printf("Chain %d\n", cas++);print_path(n, k);printf("Total distance sum = %d\n\n", f[k][n]);}return 0;}

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