本文探讨了一种将原始任务逆向转换为新的问题的方法，即给出输出，请求生成输入，使得原始问题的解决方案能够产生提供的输出。以Topcoder Open 2014 Round 2C中的Inverse RMQ任务为例。通过给出一个n×n距离矩阵，任务要求判断该矩阵是否可以代表一个有向加权树的所有权重均为正整数。详细介绍了输入输出格式、示例测试用例及解决策略，最终实现了一个高效的算法来验证矩阵是否满足有向加权树的条件。
There is an easy way to obtain a new task from an old one called "Inverse the problem": we give an output of the original task, and ask to generate an input, such that solution to the original problem will produce the output we provided. The hard task of Topcoder Open 2014 Round 2C, InverseRMQ, is a good example.
Now let's create a task this way. We will use the task: you are given a tree, please calculate the distance between any pair of its nodes. Yes, it is very easy, but the inverse version is a bit harder: you are given an n × n distance matrix. Determine if it is the distance matrix of a weighted tree (all weights must be positive integers).
The first line contains an integer n (1 ≤ n ≤ 2000) — the number of nodes in that graph.
Then next n lines each contains n integers di, j (0 ≤ di, j ≤ 109) — the distance between node i and node j.
If there exists such a tree, output "YES", otherwise output "NO".
3 0 2 7 2 0 9 7 9 0
YES
3 1 2 7 2 0 9 7 9 0
NO
3 0 2 2 7 0 9 7 9 0
NO
3 0 1 1 1 0 1 1 1 0
NO
2 0 0 0 0
NO
In the first example, the required tree exists. It has one edge between nodes 1 and 2 with weight 2, another edge between nodes 1 and 3 with weight 7.
In the second example, it is impossible because d1, 1 should be 0, but it is 1.
In the third example, it is impossible because d1, 2 should equal d2, 1.
我对于暴力出奇迹又有了更深的理解……
题意是给你一个dist[i][j]的邻接矩阵，判断这是不是一棵树。
想法是先假设这就是棵树，用最小生成树直接算出应有的n-1条边，然后暴力求出在只有这n-1条边的情况下的dist和原数组比较
当然前面还要预处理排除一堆不合法答案
这题n=2000就是400w的边，再加点处理也有200w边，明显稠密图，应该用Prim，居然Kruskal能过……服了
贴代码……
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 2147483647
#define pa pair<int,int>
#define N 2100
using namespace std;
struct bian{
int x,y,z;
}b[2000010];
bool operator < (const bian &a,const bian &b)
{
return a.z<b.z;
}
struct edge{
int to,next,v;
}e[10*N];int head[N];
LL n,cnt,tot;
LL a[N][N];
int fa[N];
int top,zhan[N];bool vis[N];
LL dist[N][N];
inline int getfa(int x)
{return fa[x]==x?x:fa[x]=getfa(fa[x]);}
inline void ins(int u,int v,int w)
{
e[++cnt].to=v;
e[cnt].next=head[u];
e[cnt].v=w;
head[u]=cnt;
}
inline void insert(int u,int v,int w)
{
ins(u,v,w);
ins(v,u,w);
}
inline LL read()
{
LL x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline void init()
{
n=read();
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
a[i][j]=read();
}
inline bool pre_judge()
{
for (int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
if (i==j&&a[i][i]!=0)return 0;
if (i!=j&&a[i][j]==0)return 0;
if (a[i][j]!=a[j][i])return 0;
}
return 1;
}
inline void Kruskal()
{
for (int i=1;i<=n;i++)fa[i]=i;
for (int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if (i<j)
{
b[++tot].x=i;
b[tot].y=j;
b[tot].z=a[i][j];
}
sort(b+1,b+tot+1);
int piece=n;
for (int i=1;i<=tot;i++)
{
int fx=getfa(b[i].x);
int fy=getfa(b[i].y);
if (fx==fy)continue;
piece--;
fa[fx]=fy;
insert(b[i].x,b[i].y,b[i].z);
if (piece==1)return;
}
}
inline void dfs(int cur)
{
for (int i=head[cur];i;i=e[i].next)
{
if (vis[e[i].to])continue;
for (int j=1;j<=top;j++)
{
dist[e[i].to][zhan[j]]=dist[zhan[j]][e[i].to]=dist[zhan[j]][cur]+e[i].v;
}
zhan[++top]=e[i].to;
vis[e[i].to]=1;
dfs(e[i].to);
}
}
int main()
{
init();
if (!pre_judge())
{
printf("NO");
return 0;
}
Kruskal();
zhan[1]=1;top=1;vis[1]=1;
dfs(1);
for (int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(a[i][j]!=dist[i][j])
{
printf("NO");
return 0;
}
printf("YES");
return 0;
}以下Prim版(第一次写，有点锉，神犇别D我)
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 2147483647
#define pa pair<int,int>
#define N 2100
using namespace std;
struct edge{
int to,next,v;
}e[10*N];int head[N];
LL n,cnt;
LL a[N][N];//读入的距离
bool inset[N];//是否在MST集合中
pa dis[N]; //二元组dist[k]=(i,j)表示从所有在集合中的点到k的最短边是从j到k,权为=i
int top,zhan[N];bool vis[N];//MST之后处理dist的dfs用
LL dist[N][N];//最后算出来的dist
inline void ins(int u,int v,int w)
{
e[++cnt].to=v;
e[cnt].next=head[u];
e[cnt].v=w;
head[u]=cnt;
}
inline void insert(int u,int v,int w)
{
ins(u,v,w);
ins(v,u,w);
}
inline LL read()
{
LL x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline bool pre_judge()
{
for (int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
if (i==j&&a[i][i]!=0)return 0;
if (i!=j&&a[i][j]==0)return 0;
if (a[i][j]!=a[j][i])return 0;
}
return 1;
}
inline void init()
{
n=read();
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
a[i][j]=read();
}
inline void prim()
{
int cur=1;inset[1]=1;
for (int i=2;i<=n;i++)
{
dis[i].first=a[1][i];
dis[i].second=1;
}
for (int i=1;i<n;i++)
{
LL mn=inf;
int from=0;
for (int j=1;j<=n;j++)
if (!inset[j]&&dis[j].first<mn)
{
mn=dis[j].first;
from=dis[j].second;
cur=j;
}
insert(from,cur,mn);
inset[cur]=1;
for (int j=1;j<=n;j++)
if (!inset[j]&&a[cur][j]<dis[j].first)
{
dis[j].first=a[cur][j];
dis[j].second=cur;
}
}
}
inline void dfs(int cur)
{
for (int i=head[cur];i;i=e[i].next)
{
if (vis[e[i].to])continue;
for (int j=1;j<=top;j++)
{
dist[e[i].to][zhan[j]]=dist[zhan[j]][e[i].to]=dist[zhan[j]][cur]+e[i].v;
}
zhan[++top]=e[i].to;
vis[e[i].to]=1;
dfs(e[i].to);
}
}
int main()
{
init();
if (!pre_judge())
{
printf("NO");
return 0;
}
prim();
zhan[1]=1;top=1;vis[1]=1;
dfs(1);
for (int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(a[i][j]!=dist[i][j])
{
printf("NO");
return 0;
}
printf("YES\n");
return 0;
}
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