codeforces472D

本文介绍了一种算法,用于判断给定的n×n最短距离矩阵是否能由一个边权皆为正整数的树所满足。通过Kruskal算法构建最小生成树并使用深度优先搜索验证距离矩阵的正确性。

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Design Tutorial: Inverse the Problem

 CodeForces - 472D 

给你了一个 n × n最短距离矩阵。(即矩阵中dis[u][v]为u点到v点的最短距离),判断是否存在一个边权皆为正整数的树,恰好满足这个最短距离矩阵 。

Input

第一行为一个整数 n (1 ≤ n ≤ 2000) — 表示图中有多少个点.

下面 n 行,每行包括 n 个整数 di, j (0 ≤ di, j ≤ 109) — 点i和点j之间的最短距离.

Output

如果存在这样的树,输出 "YES", 否则输出"NO".

Examples

Input
3
0 2 7
2 0 9
7 9 0
Output
YES
Input
3
1 2 7
2 0 9
7 9 0
Output
NO
Input
3
0 2 2
7 0 9
7 9 0
Output
NO
Input
3
0 1 1
1 0 1
1 1 0
Output
NO
Input
2
0 0
0 0
Output
NO

sol:首先很明显的性质就是在树上,两个点之间一定有且仅有一条最短的路径,于是很明显用Kruskal把MST构建出来之后暴力dfs判断距离是否相等即可
Ps:根据样例可以特判掉很多奇奇怪怪的情况,比方说Dis[x][x]!=0或者Dis[x][y]!=Dis[y][x]等等(不过对答案毫无影响)
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=2005,M=4000005;
int n,Dis[N][N];
namespace Tree
{
    int tot=0,Next[M],to[M],val[M],head[N];
    int cnt=0;
    struct Edge
    {
        int U,V,Quan;
        inline bool operator<(const Edge &tmp)const
        {
            return Quan<tmp.Quan;
        }
    }E[M];
    inline void add(int x,int y,int z)
    {
        Next[++tot]=head[x];
        to[tot]=y;
        val[tot]=z;
        head[x]=tot;
    }
    int Father[N];
    inline int GetFa(int x)
    {
        return (Father[x]==x)?(Father[x]):(Father[x]=GetFa(Father[x]));
    }
    int Root,Path[N][N];
    inline void dfs(int x,int fa,int Sum)
    {
        int i; Path[Root][x]=Sum;
        for(i=head[x];i;i=Next[i]) if(to[i]!=fa)
        {
            dfs(to[i],x,Sum+val[i]);
        }
    }
    inline void Solve()
    {
        int i,j;
        for(i=1;i<=n;i++)
        {
            Father[i]=i;
            for(j=1;j<i;j++) E[++cnt]=(Edge){j,i,Dis[j][i]};
        }
        sort(E+1,E+cnt+1);
        for(i=1;i<=cnt;i++)
        {
            Edge tmp=E[i];
            int x=tmp.U,y=tmp.V,z=tmp.Quan;
            int xx=GetFa(x),yy=GetFa(y);
            if(xx==yy) continue;
            Father[xx]=yy;
            add(x,y,z);
            add(y,x,z);
        }
        for(i=1;i<=n;i++) Root=i,dfs(i,0,0);
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++) if(Path[i][j]!=Dis[i][j])
            {
                puts("NO"); exit(0);
            }
        }
        puts("YES");
    }
}
int main()
{
    int i,j;
    R(n);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++) R(Dis[i][j]);
    }
    for(i=1;i<=n;i++)
    {
        if(Dis[i][i]) return puts("NO"),0;
        for(j=1;j<i;j++) if((Dis[j][i]!=Dis[i][j])||(!Dis[j][i])) return puts("NO"),0;
    }
    Tree::Solve();
    return 0;
}
/*
Input
3
0 2 7
2 0 9
7 9 0
Output
YES

Input
3
1 2 7
2 0 9
7 9 0
Output
NO

Input
3
0 2 2
7 0 9
7 9 0
Output
NO

Input
3
0 1 1
1 0 1
1 1 0
Output
NO

Input
2
0 0
0 0
Output
NO
*/
View Code

 

 

转载于:https://www.cnblogs.com/gaojunonly1/p/10793292.html

### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
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