[THOJ 2175] cut

Posted 2020-10-06

题意

　　已知有一张 n 个点的图, 以及两两的最小割, 构造出一张合法的原图.

　　n <= 1000 .

 

分析

　　假设存在合法原图, 最小割树即是最小生成树.

　　求出最小生成树, 判定任意点对间是否满足, 然后再输出最小割树即可.

 

实现

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cctype>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 #define F(i,a,b) for (int i=(a);i<=(b);i++)
 8 
 9 const int N=105;
10 const int L=10005;
11 const int E=205;
12 const int INF=~0u>>1;
13 
14 int n;
15 int p[N][N];
16 
17 int m;
18 struct Ed {
19     int u,v,d;
20     inline Ed(int _u=0,int _v=0,int _d=0):u(_u),v(_v),d(_d){}
21     friend inline int operator < (Ed a,Ed b) {
22         return a.d>b.d;
23     }
24 }ed[L];
25 
26 int f[N];
27 struct Graph {
28     int v,d,nx;
29     inline Graph(int _v=0,int _d=0,int _nx=0):v(_v),d(_d),nx(_nx){}
30 }mp[E];
31 int tot,hd[N];
32 
33 int dis[N][N];
34 
35 inline int rd(void) {
36     int x=0,f=1; char c=getchar();
37     for (;!isdigit(c);c=getchar()) if (c==‘-‘) f=-1;
38     for (;isdigit(c);c=getchar()) x=x*10+c-‘0‘;
39     return x*f;
40 }
41 
42 int Find(int x) {
43     if (f[x]==x) return x;
44     else return f[x]=Find(f[x]);
45 }
46 inline void Ins(int u,int v,int d) {
47     mp[++tot]=Graph(v,d,hd[u]); hd[u]=tot;
48     mp[++tot]=Graph(u,d,hd[v]); hd[v]=tot;
49 }
50 
51 void DFS(int *a,int x,int pre) {
52     for (int k=hd[x];k>0;k=mp[k].nx)
53         if (mp[k].v!=pre) {
54             a[mp[k].v]=min(a[x],mp[k].d);
55             DFS(a,mp[k].v,x);
56         }
57 }
58 
59 int main(void) {
60     #ifndef ONLINE_JUDGE
61         freopen("sdchr.in","r",stdin);
62         freopen("sdchr.out","w",stdout);
63     #endif
64     
65     cin>>n; F(i,1,n) F(j,1,n) p[i][j]=rd(); F(i,1,n) p[i][i]=INF;
66     
67     F(i,1,n) F(j,1,n)
68         ed[++m]=Ed(i,j,p[i][j]);
69     sort(ed+1,ed+m+1);
70     F(i,1,n) f[i]=i; tot=1;
71     F(i,1,m) {
72         int u=ed[i].u,v=ed[i].v,d=ed[i].d;
73         int fu=Find(u),fv=Find(v);
74         if (fu!=fv) {
75             Ins(u,v,d);
76             f[fu]=fv;
77         }
78     }
79     
80     F(i,1,n) {
81         dis[i][i]=INF;
82         DFS(dis[i],i,-1);
83     }
84     F(i,1,n) F(j,1,n) if (dis[i][j]!=p[i][j]) {
85         printf("-1\n"); return 0;
86     }
87     cout<<n-1<<endl;
88     for (int i=2;i<=tot;i+=2) {
89         int x=mp[i].v,y=mp[i^1].v,d=mp[i].d;
90         printf("%d %d %d\n",x,y,d);
91     }
92     
93     return 0;
94 }