Codeforces round #270 d design Tutorial: inverse the problem -- MST + DFS

Question: Give a distance matrix and ask if it is a correct weighted tree.

Solution: first create a minimum spanning tree based on the distance matrix, because the given distance is the shortest distance between vertices, and then run DFS on each vertex to obtain the expected dis [] [], then compare the consistency between DIS and the original MP.

First, you must judge something. For more information, see the code.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <string>#include <vector>using namespace std;#define N 2007struct Edge{    int u,v,w;}edge[N*N];int a[N][N],fa[N],dis[N][N];vector<pair<int,int> > G[N];int cmp(Edge ka,Edge kb) { return ka.w < kb.w; }int findset(int x){    if(x != fa[x])        fa[x] = findset(fa[x]);    return fa[x];}void dfs(int u,int fa,int ori){    for(int i=0;i<G[u].size();i++)    {        int v = G[u][i].first;        if(v == fa) continue;        dis[ori][v] = dis[ori][u] + G[u][i].second;        dfs(v,u,ori);    }}int main(){    int n,i,j;    while(scanf("%d",&n)!=EOF)    {        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)                scanf("%I64d",&a[i][j]);        int tag = 1;        for(i=1;i<=n;i++)        {            fa[i] = i;            for(j=1;j<=n;j++)            {                if((i == j && a[i][j] != 0)||(i != j && a[i][j] == 0)||(a[i][j] != a[j][i]))                {                    tag = 0;                    break;                }            }            if(!tag) break;        }        if(!tag) { puts("NO"); continue; }        int tot = 0;        for(i=1;i<=n;i++)            for(j=i+1;j<=n;j++)                edge[tot].u = i, edge[tot].v = j,edge[tot++].w = a[i][j];        sort(edge,edge+tot,cmp);        for(i=0;i<tot;i++)        {            int u = edge[i].u, v = edge[i].v, w = edge[i].w;            int fx = findset(u), fy = findset(v);            if(fx != fy)            {                G[u].push_back(make_pair(v,w));                G[v].push_back(make_pair(u,w));                fa[fx] = fy;            }        }        for(i=1;i<=n;i++)        {            dis[i][i] = 0;            dfs(i,0,i);            for(j=1;j<=n;j++)            {                if(a[i][j] != dis[i][j])                {                    tag = 0;                    break;                }            }            if(!tag) break;        }        if(!tag)            puts("NO");        else            puts("YES");    }    return 0;}

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Codeforces round #270 d design Tutorial: inverse the problem -- MST + DFS