题意:$n$个杯子依次编号$1$到$n$,某些杯子下面有一个小球,可以询问区间$[l,r]$中小球数量的奇偶性,花费$c_{l,r}$,问要知道所有小球的分布情况的最少花费
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询问一个区间$[l,r]$,可以看作在$l-1$和$r$之间连边
最终要使得图连通,最小花费就是最小生成树的大小
```
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define reg register
using namespace std;
typedef long long ll;
const int N=2005;
struct E
{
int from,to,dis;
inline friend bool operator < (E a,E b) {return a.dis<b.dis;}
}e[N*N];
int n,m,f[N];
inline int read()
{
int x=0,w=1;
char c=getchar();
while (!isdigit(c)&&c!='-') c=getchar();
if (c=='-') c=getchar(),w=-1;
while (isdigit(c))
{
x=(x<<1)+(x<<3)+c-'0';
c=getchar();
}
return x*w;
}
int find(int x){return f[x]==x?x:f[x]=find(f[x]);}
inline ll Kruskal(ll res=0)
{
sort(e+1,e+m+1);
for (reg int i=1;i<=n;i++) f[i]=i;
for (reg int i=1,cnt=0;i<=m;i++)
{
int u=find(e[i].from),v=find(e[i].to);
if (u==v) continue;
f[u]=v; res+=e[i].dis;
}
return res;
}
int main()
{
n=read();
for (reg int i=1;i<=n;i++)
for (reg int j=i;j<=n;j++) e[++m]=(E){i-1,j,read()};
printf("%lld\n",Kruskal());
return 0;
}
```