Poj 1679 The Unique MST(次小生成树)

这道题题意是：给你一个图，然后判断这个图形成的最小生成树是否唯一？

解法：可以用求次小生成树解决，求出次小生成树和最小生成树的权值比较，相等就输出不唯一，否则，输出最小生成树的权值。

#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <string.h>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
#include <iostream>
#define PI 3.1415926535898
#define LL long long
#define MAX 0x3fffffff
#define INF 0x3f3f3f3f
#define mem(a,v) memset(a,v,sizeof(a))
const int MAX_N = 1e5+10;
const double eps = 1e-7;
const int mod = 10007;
const LL inf = 1LL<<60;
using namespace std;
int cost[200][200];
int n,m;
int used[200][200];
int Ma[200][200];
int vis[200];
int pre[200];
int d[200];
int prim()
{
int sum = 0;
memset(vis,0,sizeof(vis));
memset(used,0,sizeof(used));
memset(Ma,0,sizeof(Ma));
d[0] = 0;
vis[0] = 1;
pre[0] = -1;
for(int i = 1;i < n;i++)
{
d[i] = cost[0][i];
pre[i] = 0;
}
for(int i = 1;i < n;i++)
{
int Mi = INF;
int p = -1;
for(int j = 0;j < n;j++)
{
if(!vis[j]&&d[j] < Mi)
{
Mi = d[j];
p = j;
}
}
if(Mi == INF) return -1;
sum+=Mi;
vis[p] = 1;
used[p][pre[p]] = used[pre[p]][p] = 1;
for(int j = 0;j < n;j++)
{
if(vis[j])
Ma[j][p] = Ma[p][j] = max(Ma[j][pre[p]],d[p]);
if(!vis[j]&&d[j] > cost[p][j])
{
d[j] = cost[p][j];
pre[j] = p;
}
}
}
return sum;
}
int Mst;
int judge()
{
int cMst = INF;
for(int i = 0;i < n;i++)
{
for(int j = i+1;j < n;j++)
{
if(cost[i][j]!=INF&&!used[i][j])
cMst = min(cMst,Mst+cost[i][j]-Ma[i][j]);
}
}
if(cMst == INF) return -1;
return cMst;
}
int main()
{
//freopen("out.txt","w",stdout);
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
int u,v,w;
for(int i = 0;i < n;i++)
{
for(int j = 0;j < n;j++)
cost[i][j] = INF;
cost[i][i] = 0;
}
for(int i = 0;i < m;i++)
{
scanf("%d%d%d",&u,&v,&w);
cost[v-1][u-1] = cost[u-1][v-1] = w;
}
Mst = prim();
if(Mst == -1||Mst == judge())
printf("Not Unique!\n");
else
printf("%d\n",Mst);
}
return 0;
}
也可以用Kruskal解决，先把边权相等的边标记，然后先求一次Kruskal，再把刚才求的最小生成树上的边标记，然后去掉两个都有的边再求一次最小生成树，看这两个求的值是否相同，相同就输出不唯一，否则，输出答案。
#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <string.h>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
#include <iostream>
#define PI 3.1415926535898
#define LL long long
#define MAX 0x3fffffff
#define INF 0x3f3f3f3f
#define mem(a,v) memset(a,v,sizeof(a))
const int MAX_N = 1e5+10;
const double eps = 1e-7;
const int mod = 10007;
const LL inf = 1LL<<60;
using namespace std;
struct node
{
int u,v,w,flag,x;
}es[MAX_N];
int n,m,fa[MAX_N];
int vis[MAX_N];
bool cmp(const node &x,const node &y)
{
return x.w < y.w;
}
void init()
{
for(int i = 0;i <= MAX_N;i++)
fa[i] = i;
}
int Find(int x)
{
return x == fa[x]?fa[x]:fa[x] = Find(fa[x]);
}
void mix(int x,int y)
{
int fx = Find(x);
int fy = Find(y);
if(fx!=fy)
fa[fy] = fx;
}
int tmp;
int Kruskal()
{
init();
int sum = 0;
int cnt = 0;
for(int i = 0;i < m;i++)
{
if(es[i].x) continue;
if(Find(es[i].u)!=Find(es[i].v))
{
sum+=es[i].w;
cnt++;
mix(es[i].u,es[i].v);
if(!tmp) es[i].flag = 1;
if(cnt == n-1)
break;
}
}
return sum;
}
int main()
{
//freopen("out.txt","w",stdout);
int t;
scanf("%d",&t);
while(t--)
{
memset(vis,0,sizeof(vis));
scanf("%d%d",&n,&m);
int u,v,w;
for(int i = 0;i < m;i++)
{
scanf("%d%d%d",&u,&v,&w);
es[i].u = u; es[i].v = v; es[i].w = w;
es[i].flag = 0;
es[i]
4000
.x = 0;
}
sort(es,es+m,cmp);
for(int i = 1;i < m;i++)
if(es[i-1].w == es[i].w)
vis[i-1] = vis[i] = 1;
tmp = 0;
int ans1 = Kruskal(),fg = 0;
tmp = 1;
for(int i = 0;i < m;i++)
{
if(vis[i]&&es[i].flag)
{
es[i].x = 1;
int ans2 = Kruskal();
if(ans1 == ans2)
{
fg = 1;
printf("Not Unique!\n");
break;
}
es[i].x = 0;
}
}
if(!fg)
printf("%d\n",ans1);
}
return 0;
}