设d(i, j)为连通第i个点到第j个点的树的最小长度，则有状态转移方程：
d(i, j) = min{ d(i, k) + d(k + 1, j) + p[k].y - p[j].y + p[k+1].x - p[i].x }
然后用四边形不等式优化之。。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <map> 6 #define MP make_pair 7 #define x first 8 #define y second 9 using namespace std; 10 11 typedef pair<int, int> PII; 12 13 const int maxn = 1000 + 10; 14 const int INF = 0x3f3f3f3f; 15 int n; 16 17 PII p[maxn]; 18 19 int d[maxn][maxn], s[maxn][maxn]; 20 21 int main() 22 { 23 while(scanf("%d", &n) == 1 && n) 24 { 25 for(int i = 1; i <= n; i++) scanf("%d%d", &p[i].x, &p[i].y); 26 for(int i = 1; i <= n; i++) 27 { 28 s[i][i + 1] = i; 29 d[i][i + 1] = p[i+1].x - p[i].x + p[i].y - p[i+1].y; 30 } 31 32 for(int l = 3; l <= n; l++) 33 { 34 for(int i = 1; i + l - 1 <= n; i++) 35 { 36 int j = i + l - 1; 37 d[i][j] = INF; 38 for(int k = s[i][j-1]; k <= s[i+1][j]; k++) 39 { 40 int t = d[i][k] + d[k+1][j] + p[k].y - p[j].y + p[k+1].x - p[i].x; 41 if(t < d[i][j]) 42 { 43 d[i][j] = t; 44 s[i][j] = k; 45 } 46 } 47 } 48 } 49 50 printf("%d ", d[1][n]); 51 } 52 53 return 0; 54 }