UVA662- Fast Food

题意：在一条公路上，有n个酒店，要建造k个供给站(建造在酒店所在的位置)，给出酒店的位置，求怎么样建造供给站才干使得每一个酒店都能得到服务且所要走的路程最短。

思路：在i到j酒店建立一个供给站，要使得路程和最短，要将供给站建立在中间。假设i到j为偶数时，能够建立在中间两个数当中一个地方，假设是奇数时，应该建立在(i + j) / 2的地方。我们能够预处理从i到j酒店的最短路程和dis[i][j]。所以能够得到d[i][j]表示i个供给站为前j个酒店服务时的最短路程和。之后的输出能够在计算的时候增加一个记录数组，记录切割的地方，然后递归输出。

状态转移方程：d[i][j] = min(d[i][j], d[i - 1][k] + dis[k + 1][j]); (1 <= k <  j)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

typedef long long ll;
using namespace std;

const int MAXN = 205;
const int INF = 0x3f3f3f3f;

int p[MAXN], vis[MAXN][MAXN]; 
ll dis[MAXN][MAXN], d[MAXN][MAXN];
int n, m;

void init() {
    memset(vis, 0, sizeof(vis));
    memset(dis, 0, sizeof(dis));
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            for (int k = i; k <= j; k++)
                dis[i][j] += abs(p[(i + j) / 2] - p[k]);

    for (int i = 0; i <= m; i++)
        for (int j = 0; j <= n; j++)
            d[i][j] = INF;

    for (int i = 1; i <= n; i++)
        d[1][i] = dis[1][i];
}

void outPut(int cnt, int x, int y) {
    printf("Depot %d at restaurant %d serves ", cnt, (x + y) / 2);
    if (x == y)
        printf("restaurant %d
", x);
    else 
        printf("restaurants %d to %d
", x, y);
}

void print(int x, int y) {
    if (x > 1)
        print(x - 1, vis[x][y]);
    outPut(x, vis[x][y] + 1, y); 
}

int main() {
    int t = 1;
    while (scanf("%d %d", &n, &m) && n || m) {
        for (int i = 1; i <= n; i++) 
            scanf("%d", &p[i]);
        init();

        for (int i = 2; i <= m; i++)
            for (int j = i; j <= n; j++) 
                for (int k = i - 1; k < j; k++) 
                    if (d[i][j] > d[i - 1][k] + dis[k + 1][j]) {
                        d[i][j] = d[i - 1][k] + dis[k + 1][j]; 
                        vis[i][j] = k;
                    }

        printf("Chain %d
", t++); 
        print(m, n);
        printf("Total distance sum = %lld

", d[m][n]);

    }    
}