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这题一看就像是动归，但我就是想不出来怎么动归。。。
纠结了n天之后，果断百度，果然是动归
a[i][j]表示前i个饭店如果需要j个depot
c[i][j]表示如果从第i个到第j个饭店都由一个depot负责，的最短距离
于是就有了神奇的动态转移方程
a[i][j] = min{a[i][j], a[i-k][j-1]+c[i-k+1][i]} (k = 1, 2, …, i-1)
嗯，然后就益母聊染了
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 5 #define min(a,b) (((a)>(b))?(b):(a)) 6 7 #define MAXN 300 8 9 int d[MAXN];//distance between restautants i and headquarter 10 int a[MAXN][MAXN];//the lowest distance while the first i restautants need j depots 11 int c[MAXN][MAXN];//the lowest distance while restautant i to restaurant j need 1 depots 12 13 int main(void) 14 { 15 int n, k; 16 int num = 1; 17 scanf("%d%d", &n, &k); 18 while(n > 0 && k > 0){ 19 for(int i = 1; i <= n; ++i) scanf("%d", &d[i]); 20 memset(c, 0, sizeof(c)); 21 for(int i = 1; i <= n; ++i){ 22 for(int j = i+1; j <= n; ++j) 23 for(int k = i, p = (i+j)/2; k <= j; ++k) 24 c[i][j] += abs(d[k] - d[p]); 25 } 26 for(int i = 0; i <= n; ++i) 27 for(int j = 0; j <= n; ++j) 28 a[i][j] = 0xFFFFFFF; 29 a[1][1] = 0; 30 for(int i = 2; i <= n; ++i){ 31 a[i][1] = c[1][i]; 32 for(int j = 2; j <= i && j <= k; ++j) 33 for(int k = 1; k < i; ++k) 34 if(a[i][j] > a[i-k][j-1]+c[i-k+1][i]){ 35 a[i][j] = a[i-k][j-1]+c[i-k+1][i]; 36 /*for(int i = 0; i < n; ++i) printf("%12d", i); 37 printf("\n"); 38 for(int i = 0; i <= n; ++i){ 39 printf("%2d: ", i); 40 for(int j = 0; j <= n; ++j) printf("%12d", a[i][j]); 41 printf("\n"); 42 }*/ 43 //printf("a[%d][%d] = a[%d-%d][%d-1] + c[%d+1][%d]\n", i, j, i, k, j, k, i); 44 //printf("%d = %d + %d\n", a[i][j], a[i-k][j-1], c[k+1][i]); 45 } 46 } 47 printf("Chain %d\n", num++); 48 printf("Total distance sum = %d\n\n", a[n][k]); 49 50 scanf("%d%d", &n, &k); 51 } 52 return 0; 53 }
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原文地址：http://www.cnblogs.com/xuezhonghao/p/4926379.html
