版权声明:转载请声明出处,谢谢配合。 https://blog.csdn.net/zxyoi_dreamer/article/details/82707219
参考题目:POJ2728
解析:
就是套用01分数规划的两种方法,然后根据题目和数据规模选择用 还是 算法求最大还是最小生成树就行了。
显然这道题适合用 求最小生成树,毕竟是完全图。
代码(二分):
#include<cstring>
#include<queue>
#include<cctype>
#include<algorithm>
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
#define ll long long
#define re register
#define gc getchar
#define pc putchar
#define cs const
#define st static
inline
int getint(){
re int num=0;
re bool f=1;
re char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=0;ch=gc();}
while(isdigit(ch))num=(num<<3)+(num<<1)+(ch^48),ch=gc();
return f?num:-num;
}
int x[1005],y[1005],h[1005];
int n;
double cost[1005][1005];
double dist[1005][1005];
double w[1005];
bool vis[1005];
inline
bool prim(double k){
double mint=0.0;
for(int re i=1;i<=n;++i){
w[i]=cost[1][i]-dist[1][i]*k;
}
memset(vis,0,sizeof vis);
vis[1]=true;
for(int re kk=2;kk<=n;++kk){
double minn=1e9;
int pos=0;
for(int re i=1;i<=n;++i){
if(vis[i])continue;
if(w[i]<minn)minn=w[i],pos=i;
}
vis[pos]=1;
mint+=minn;
for(int re i=1;i<=n;++i){
if(vis[i])continue;
if(w[i]>cost[pos][i]-dist[pos][i]*k){
w[i]=cost[pos][i]-dist[pos][i]*k;
}
}
}
return mint>=0.0;
}
signed main(){
while(n=getint()){
for(int re i=1;i<=n;++i)x[i]=getint(),y[i]=getint(),h[i]=getint();
for(int re i=1;i<n;++i){
for(int re j=i+1;j<=n;++j)
dist[i][j]=dist[j][i]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])),
cost[i][j]=cost[j][i]=fabs(h[i]-h[j]);
}
double L=0.0,R=100.0;
while(R-L>1e-6){
double mid=(L+R)/2;
if(prim(mid))L=mid;
else R=mid;
}
printf("%.3f\n",L);
}
return 0;
}
代码( ):
#include<cstring>
#include<queue>
#include<cctype>
#include<algorithm>
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
#define ll long long
#define re register
#define gc getchar
#define pc putchar
#define cs const
#define st static
inline
int getint(){
re int num=0;
re bool f=1;
re char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=0;ch=gc();}
while(isdigit(ch))num=(num<<3)+(num<<1)+(ch^48),ch=gc();
return f?num:-num;
}
int x[1005],y[1005],h[1005];
int n;
double cost[1005][1005];
double dist[1005][1005];
double w[1005];
bool vis[1005];
double val[1005];
double ans,L;
inline
double prim(double k){
double x=0.0,y=0.0;
for(int re i=1;i<=n;++i){
w[i]=cost[1][i]-dist[1][i]*k;
val[i]=cost[1][i];
}
memset(vis,0,sizeof vis);
vis[1]=true;
for(int re kk=2;kk<=n;++kk){
double minn=1e9;
int pos=0;
for(int re i=1;i<=n;++i){
if(vis[i])continue;
if(w[i]<minn)minn=w[i],pos=i;
}
vis[pos]=1;
x+=val[pos];
y+=val[pos]-w[pos];
for(int re i=1;i<=n;++i){
if(vis[i])continue;
if(w[i]>cost[pos][i]-dist[pos][i]*k){
w[i]=cost[pos][i]-dist[pos][i]*k;
val[i]=cost[pos][i];
}
}
}
return x*k/y;
}
signed main(){
while(n=getint()){
double maxn=-1e9,minn=1e9;
for(int re i=1;i<=n;++i)x[i]=getint(),y[i]=getint(),h[i]=getint();
for(int re i=1;i<=n;++i){
for(int re j=i+1;j<=n;++j)
dist[i][j]=dist[j][i]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])),
cost[i][j]=cost[j][i]=fabs(h[i]-h[j]),
minn=min(minn,cost[i][j]),
maxn=max(maxn,dist[i][j]);
}
L=minn/maxn;
while(true){
ans=prim(L);
if(fabs(ans-L)<1e-8)break;
L=ans;
}
printf("%.3f\n",ans);
// printf("%.3lf\n",ans);这样写有毒,会WA
}
return 0;
}