DP优化/四边形不等式
这题……跟邮局那题简直一模一样吧……好水的E题……
设dp[i][j]表示前 i 艘“gondola”坐了前 j 个人，那么方程即为$dp(i,j)=min\{ dp[i-1][k]+w[k][j] \} (i\leq k\leq j)$
很明显$w(l,r)=\sum_{i=l}^r \sum_{j=l}^r u(i,j) /2$是满足四边形不等式的……那么根据决策单调性直接搞就行了……
//CF 321E
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define rep(i,n) for(int i=0;i<n;++i)
#define F(i,j,n) for(int i=j;i<=n;++i)
#define D(i,j,n) for(int i=j;i>=n;--i)
using namespace std;
//#define debug
int getint(){
int v=,sign=; char ch=getchar();
while(ch<''||ch>'') {if (ch=='-') sign=-; ch=getchar();}
while(ch>=''&&ch<='') {v=v*+ch-''; ch=getchar();}
return v*sign;
}
typedef long long LL;
const int N=,INF=~0u>>;
/*******************tamplate********************/
LL u[][],n,m,w[][],dp[][];
int s[][];
int main(){
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
// freopen("output.txt","w",stdout);
#endif
n=getint(); m=getint();
F(i,,n) F(j,,n) u[i][j]=getint();
F(i,,n) F(j,,n) u[i][j]+=u[i-][j]+u[i][j-]-u[i-][j-]; F(i,,n) F(j,i+,n) w[i][j]=u[j][j]-u[i-][j]-u[j][i-]+u[i-][i-]; #ifdef debug
F(i,,n) {F(j,,n) printf("%3d",w[i][j]);puts("");}
#endif
F(i,,m) F(j,,n) dp[i][j]=INF;
F(i,,n){
dp[][i]=w[][i];
s[][i]=;
}
F(i,,m){
s[i][n+]=n;
D(j,n,i)
F(k,s[i-][j],s[i][j+])
if (dp[i-][k]+w[k+][j]<dp[i][j]){
s[i][j]=k;
dp[i][j]=dp[i-][k]+w[k+][j];
}
}
printf("%I64d\n",dp[m][n]/);
return ;
}