poj 1679 The Unique MST【次小生成树】
The Unique MST
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24034 | Accepted: 8535 |
Description
Given a connected undirected graph, tell if its minimum spanning tree is unique.
Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the following properties:
1. V' = V.
2. T is connected and acyclic.
Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E') of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E'.
Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the following properties:
1. V' = V.
2. T is connected and acyclic.
Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E') of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E'.
Input
The first line contains a single integer t (1 <= t <= 20), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m (1 <= n <= 100), the number of nodes and edges. Each of the following m lines contains a triple (xi, yi, wi), indicating that xi and yi are connected by an edge with weight = wi. For any two nodes, there is at most one edge connecting them.
Output
For each input, if the MST is unique, print the total cost of it, or otherwise print the string 'Not Unique!'.
Sample Input
2 3 3 1 2 1 2 3 2 3 1 3 4 4 1 2 2 2 3 2 3 4 2 4 1 2
Sample Output
3 Not Unique!
今天看了一天的次小生成树，还是有几个细节不是太理解，今晚再跟舍友探讨一下
题意：判断是不是有唯一的最小生成树（即连接所有点的边的权值之和是最小且唯一的）如果是则输出最小生成树（mst）的权值，否则输出 Not Unique！
题解：先利用prime算法求出最小生成树，在prime求MST的过程中 用数组存储MST里面任意两点间的唯一的路中 权值最大的那条边的权值。
最后枚举不在MST里面的边<i,，j>，判断<i，j>的权值 是否 和 MST里面 i 到 j 的最大权值相等，只要有一条边满足就可以说明MST不唯一。
（因为我们可以用这条不在MST的边来 代替 在MST的边，这样MST肯定不唯一）
MST更新：MST[ next ][ j ] = max(MST[ set[ next ] ][ j ], low[ next ])。（ j 属于MST里面的点）
#include<stdio.h>
#include<string.h>
#define MAX 110
#define max(x,y)(x>y?x:y)
#define INF 0x3f3f3f3f
int map[MAX][MAX],vis[MAX],set[MAX];
int low[MAX];
int mst[MAX][MAX];//记录 mst中i到j的最大权值
int inmst[MAX][MAX];//判断边是否在mst中
int n,m;
void init()//初始化
{
int i,j;
for(i=0;i<=n;i++)
{
for(j=0;j<=n;j++)
{
if(i==j)
map[i][j]=0;
else
map[i][j]=INF;
}
}
}
void getmap()//建图
{
int i,a,b,c;
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&a,&b,&c);
if(map[a][b]>c)
map[a][b]=map[b][a]=c;
}
}
void prim()
{
int i,j,min,mindis=0,next;
memset(mst,0,sizeof(mst));
memset(inmst,0,sizeof(inmst));
for(i=1;i<=n;i++)
{
low[i]=map[1][i];
vis[i]=0;
set[i]=1;//所有点的前驱都是起点
}
vis[1]=1;
for(i=1;i<n;i++)
{
min=INF;
for(j=1;j<=n;j++)
{
if(!vis[j]&&min>low[j])
{
min=low[j];
next=j;
}
}
mindis+=min;
vis[next]=1;
int fa=set[next];//求出当前新起点的前驱
inmst[fa][next]=inmst[next][fa]=1;
for(j=1;j<=n;j++)
{
if(vis[j]&&j!=next)
mst[j][next]=mst[next][j]=max(mst[fa][j],low[next]);
if(!vis[j]&&low[j]>map[next][j])
{
low[j]=map[next][j];
set[j]=next;//以新起点为所有点的新前驱
}
}
}
for(i=1;i<=n;i++)
{
for(j=1;j<i;j++)
{
if(map[i][j]!=INF&&!inmst[i][j])//不在最小生成树中且存在的边
{
if(map[i][j]==mst[i][j])//代表可以有边替代最小生成树中的边（最小生成树不唯一）
{
printf("Not Unique!\n");
return ;
}
}
}
}
printf("%d\n",mindis);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
init();
getmap();
prim();
}
return 0;
}
