[CF773D]Perishable Roads
[CF773D]Perishable Roads
题目大意:
一个\(n(n\le2000)\)个点的完全图\(G\),定义\(d(x)\)为生成树上点\(x\)到根路径上的最小边权。问图\(G\)的生成树\(\sum d(x)\)最小是多少?
思路:
由题解得到图的一些性质,然后就不难了。
源代码:
#include<cstdio>
#include<cctype>
#include<climits>
#include<algorithm>
using int64=long long;
inline int getint() {
register char ch;
while(!isdigit(ch=getchar()));
register int x=ch^'0';
while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
return x;
}
constexpr int N=2001;
bool vis[N];
int w[N][N],d[N];
int main() {
const int n=getint();
int min=INT_MAX;
for(register int i=1;i<=n;i++) {
for(register int j=i+1;j<=n;j++) {
min=std::min(min,w[i][j]=w[j][i]=getint());
}
}
d[0]=INT_MAX;
for(register int i=1;i<=n;i++) {
d[i]=INT_MAX;
for(register int j=1;j<=n;j++) {
if(i==j) continue;
w[i][j]-=min;
d[i]=std::min(d[i],w[i][j]*2);
}
}
for(register int i=1;i<=n;i++) {
int k=0;
for(register int j=1;j<=n;j++) {
if(!vis[j]&&d[j]<d[k]) k=j;
}
vis[k]=true;
for(register int j=1;j<=n;j++) {
d[j]=std::min(d[j],d[k]+w[k][j]);
}
}
for(register int i=1;i<=n;i++) {
printf("%lld\n",(int64)min*(n-1)+d[i]);
}
return 0;
}
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