description

給出有向圖的點數(shù)\(n\)和鄰接矩陣\(G\),
求\[P=∑_{1≤x,y,z≤n,x≠y,y≠z}d(x,y,z)\]
其中\(d(x,y,z)\)表示從\(x\)不經(jīng)過\(y\)到\(z\)的最短路,如果無法到達則為\(-1\)

data range

\[4≤n≤300,?1≤G_{i,j}≤10000,G_{i,i}=0\]

solution

首先你要知道\(floyed\)的本質(zhì)是枚舉中轉(zhuǎn)點。

那么不經(jīng)過\(y\)點的最短路矩陣相當(dāng)于使用除\(y\)以外的點作為中轉(zhuǎn)點更新后的鄰接矩陣

于是考慮分治,遞歸到\(l==r\)時僅有節(jié)點\(l\)未被作為中轉(zhuǎn)點
那么在做到\([l,r]\)時，使用\([l,mid]\)更新\([mid+1,r]\)，使用\([mid+1,r]\)更新\([l,mid]\)即可。

Code

#include<bits/stdc++.h> #include<algorithm> #include<iostream> #include<cstdlib> #include<iomanip> #include<cstring> #include<complex> #include<vector> #include<cstdio> #include<string> #include<bitset> #include<ctime> #include<cmath> #include<queue> #include<stack> #include<map> #include<set> #define Cpy(x,y) memcpy(x,y,sizeof(x)) #define Set(x,y) memset(x,y,sizeof(x)) #define FILE "a" #define mp make_pair #define pb push_back #define RG register #define il inline using namespace std; typedef unsigned long long ull; typedef vector<int>VI; typedef long long ll; typedef double dd; const int N=200010; const int M=1000010; const dd eps=1e-5; const int inf=2147483647; const ll INF=1ll<<60; const ll P=100000; il ll read(){RG ll data=0,w=1;RG char ch=getchar();while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();if(ch=='-')w=-1,ch=getchar();while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();return data*w; }il void file(){srand(time(NULL)+rand());freopen(FILE".in","r",stdin);freopen(FILE".out","w",stdout); }int n;ll ans,g[20][305][305]; void divide(int l,int r,int d){if(l==r){for(RG int i=1;i<=n;i++)for(RG int j=1;j<=n;j++)if(i!=l&&j!=l)ans+=g[d-1][i][j];return;}RG int mid=(l+r)>>1;memcpy(g[d],g[d-1],sizeof(g[d]));for(RG int k=mid+1;k<=r;k++)for(RG int i=1;i<=n;i++)if(k!=i)for(RG int j=1;j<=n;j++)if(k!=j&&i!=j&&g[d][i][k]!=-1&&g[d][k][j]!=-1)if(g[d][i][j]==-1||g[d][i][j]>g[d][i][k]+g[d][k][j])g[d][i][j]=g[d][i][k]+g[d][k][j];divide(l,mid,d+1);memcpy(g[d],g[d-1],sizeof(g[d]));for(RG int k=l;k<=mid;k++)for(RG int i=1;i<=n;i++)if(k!=i)for(RG int j=1;j<=n;j++)if(k!=j&&i!=j&&g[d][i][k]!=-1&&g[d][k][j]!=-1)if(g[d][i][j]==-1||g[d][i][j]>g[d][i][k]+g[d][k][j])g[d][i][j]=g[d][i][k]+g[d][k][j];divide(mid+1,r,d+1); }int main() {n=read();for(RG int i=1;i<=n;i++)for(RG int j=1;j<=n;j++)g[0][i][j]=read();divide(1,n,1);printf("%lld\n",ans);return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/cjfdf/p/9703587.html

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