和石子合并那题类似。。。用四边形优化即可。。。
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 1005
#define maxm 300006
#define eps 1e-10
#define mod 1315423911
#define INF 1e9
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
struct node
{
int x, y;
}p[maxn];
int dp[maxn][maxn];
int s[maxn][maxn];
int n;
void read(void)
{
for(int i = 1; i <= n; i++) scanf("%d%d", &p[i].x, &p[i].y);
}
void work(void)
{
for(int i = 0; i <= n; i++)
for(int j = 0; j <= n; j++)
dp[i][j] = INF;
for(int i = 1; i <= n; i++) dp[i][i] = 0, s[i][i] = i;
for(int len = 2; len <= n; len++) {
for(int i = 1; i + len -1 <= n; i++) {
int j = i + len - 1, kk;
int a = s[i][j-1], b = min(s[i+1][j], j-1);
for(int k = a; k <= b; k++) {
int w = p[k+1].x - p[i].x + p[k].y - p[j].y;
if(dp[i][j] > dp[i][k] + dp[k+1][j] + w)
kk = k, dp[i][j] = dp[i][k] + dp[k+1][j] + w;
}
s[i][j] = kk;
}
}
printf("%d\n", dp[1][n]);
}
int main(void)
{
while(scanf("%d", &n)!=EOF) {
read();
work();
}
return 0;
}
