BZOJ1232 [Usaco2008Nov]安慰奶牛cheer
题面在这里
题目意思不清楚……其实是DFS遍历的意思
这样每条边被算了2次，点被算了度数次，起点再加一次
考虑每条边对答案的贡献，就是端点点权+边权*2
然后取最小点权作为起点就好了
示例程序：
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| #include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
inline char nc(){
static char buf[100000],*l=buf,*r=buf;
return l==r&&(r=(l=buf)+fread(buf,1,100000,stdin),l==r)?EOF:*l++;
}
inline int red(){
int res=0,f=1;char ch=nc();
while (ch<'0'||'9'<ch) {if (ch=='-') f=-f;ch=nc();}
while ('0'<=ch&&ch<='9') res=res*10+ch-48,ch=nc();
return res*f;
}
const int maxn=10005,maxe=100005;
int n,e,c[maxn],f[maxn],fa[maxn];ll ans;
inline int getfa(int x) {return fa[x]==x?x:fa[x]=getfa(fa[x]);}
struct edge{
int x,y,w;
bool operator<(const edge&b)const {return w<b.w;}
}a[maxe];
void kruskal(){
for (int i=1;i<=n;i++) fa[i]=i;
for (int i=1;i<=e;i++){
if (getfa(a[i].x)==getfa(a[i].y)) continue;
fa[getfa(a[i].x)]=getfa(a[i].y);
ans+=a[i].w*2;f[a[i].x]++;f[a[i].y]++;
}
}
int main(){
n=red(),e=red();int mn=0x3f3f3f3f;
for (int i=1;i<=n;i++) c[i]=red(),mn=min(mn,c[i]);
for (int i=1;i<=e;i++) a[i].x=red(),a[i].y=red(),a[i].w=red()*2+c[a[i].x]+c[a[i].y];
sort(a+1,a+1+e);
for (int i=1;i<=n;i++) fa[i]=i;
for (int i=1;i<=e;i++){
if (getfa(a[i].x)==getfa(a[i].y)) continue;
fa[getfa(a[i].x)]=getfa(a[i].y);ans+=a[i].w;
}
printf("%lld",ans+mn);
return 0;
}
|
