uva 10600 ACM Contest and Blackout

题目：ACM Contest and Blackout

题意：给出一张无向图，求最小生成树及次小生成树（非严格）。

思路：求次小生成树。

参考

代码：#include<bits/stdc++.h>
using namespace std;

#define maxn 5000
#define maxm 200000

struct Edge {
int x,y,z;
Edge(int xx=0,int yy=0,int zz=0) {
x=xx,y=yy,z=zz;
}
bool operator < (const Edge& oth) const {
return z<oth.z||(z==oth.z&&(x<oth.x||(x==oth.x&&y<oth.y)));
}
};

int n,m;
Edge a[maxm+5];
int fa[maxm+5];

vector<Edge> tr[maxn+5];
bool intree[maxn+5];

int d[maxn+5][maxn+5];
bool isuse[maxn+5];

void init() {
for(int i=1; i<=maxm; i++) fa[i]=i;
memset(d,0,sizeof(d));
memset(isuse,0,sizeof(isuse));
for(int i=1;i<=maxn;i++) tr[i].clear();
memset(intree,0,sizeof(intree));
}

void readin() {
scanf("%d%d",&n,&m);
for(int i=1; i<=m; i++) {
scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
}
}

int find(int x) {
return fa[x]==x?x:fa[x]=find(fa[x]);
}

int kruskal() {
int ans=0,cnt=0;
for(int i=1; i<=m; i++) {
int fa1=find(a[i].x),fa2=find(a[i].y);
if(fa1==fa2) continue;
fa[fa1]=fa2;
ans+=a[i].z;
cnt++;

tr[a[i].x].push_back(a[i]);
tr[a[i].y].push_back(Edge(a[i].y,a[i].x,a[i].z));
intree[i]=true;

if(cnt==n-1) break;
}
return ans;
}

void dfs(int x,int fa,int y) {
for(int i=1;i<=n;i++){
if(isuse[i]) d[x][i]=d[i][x]=max(d[i][fa],y);
}
isuse[x]=true;
for(int i=0;i<tr[x].size();i++){
int v=tr[x][i].y;
if(v!=fa) dfs(v,x,tr[x][i].z);
}
}

int slv(){
int ans=(1<<30);
for(int i=1;i<=m;i++){
if(!intree[i]) ans=min(ans,-d[a[i].x][a[i].y]+a[i].z);
}
return ans;
}

int main() {
int T;
scanf("%d",&T);
while(T--) {
init();
readin();
sort(a+1,a+m+1);
int ans1=kruskal();
dfs(1,0,0);
int ans2=slv();
printf("%d %d\n",ans1,ans1+ans2);
}
return 0;
}