jskA1108(flyod+分治)

题目链接

https://nanti.jisuanke.com/t/A1108

题解

我们只能处理经过连续几个点的最短路,然后加一个点的复杂度为 $O(n^2)$ ,而如果是两块合并的复杂度是 $O(n^3)$ ,这个是我们要避免的。然后就要尽量使用单点加入,这个和背包就有点类似了,所以仿照分治背包进行分治,分治到当前区间表示当前区间不能经过的最短路,然后只要统计叶子的最短路就可以了。。




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<complex>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define mid (x+y>>1)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define NM 305
#define nm 20005
using namespace std;
const double pi=acos(-1);
const int inf=1e9+7;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}



int n,d[20][NM][NM];
ll ans;

void div(int k,int x,int y){
if(x==y){
inc(i,1,n)if(i!=x)inc(j,1,n)if(j!=x){if(d[k][i][j]<inf)ans+=d[k][i][j];else ans--;}
return;
}
inc(i,1,n)inc(j,1,n)d[k+1][i][j]=d[k][i][j];
inc(v,mid+1,y)inc(i,1,n)inc(j,1,n)d[k+1][i][j]=min(d[k+1][i][j],d[k+1][i][v]+d[k+1][v][j]);
div(k+1,x,mid);
inc(i,1,n)inc(j,1,n)d[k+1][i][j]=d[k][i][j];
inc(v,x,mid)inc(i,1,n)inc(j,1,n)d[k+1][i][j]=min(d[k+1][i][j],d[k+1][i][v]+d[k+1][v][j]);
div(k+1,mid+1,y);
}

int main(){
n=read();
inc(i,1,n)inc(j,1,n){
int t=read();if(t==-1)t=inf;
d[0][i][j]=t;
}
div(0,1,n);
return 0*printf("%lld\n",ans);
}