AW348 沙漠之王（0/1分数规划-Dinkelbach）

题目地址

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易错点：

• double类型不初始化为0就爆炸.
• 每次prim前需要先从首都向每个村庄加边.
• "长度"指二维欧氏距离.

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#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN=2e3,INF=1<<30;
struct village{
	int x,y,z;
}villages[MAXN];
double getDis(village a,village b){
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.z-b.z)*(a.z-b.z));
}
int n,pre[MAXN];
bool vis[MAXN];
double minw[MAXN],cost[MAXN][MAXN],len[MAXN][MAXN];
double prim(double k){
	memset(vis,0,sizeof(vis));
	vis[1]=1;
	for(int i=2;i<=n;i++){
		minw[i]=cost[1][i]-len[1][i]*k;
		pre[i]=1;
	}
	double Cost,Len;
	Cost=Len=0;//不初始化就爆炸 
	for(int i=1;i<=n;i++){
		double nowMin=INF;
		int nowV=0;
		for(int j=1;j<=n;j++){
			if(!vis[j]&&minw[j]<nowMin){
				nowMin=minw[nowV=j];
			}
		}
		vis[nowV]=1;
		Cost+=cost[pre[nowV]][nowV],Len+=len[pre[nowV]][nowV];
		for(int j=1;j<=n;j++){
			double tmp=cost[nowV][j]-k*len[nowV][j];
			if(!vis[j]&&tmp<minw[j]){
				minw[j]=tmp;
				pre[j]=nowV;
			}
		}
	}
	return Cost/Len;
}
void init(){
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			cost[i][j]=fabs(villages[i].y-villages[j].y);
			len[i][j]=getDis(villages[i],villages[j]);
		}
	}
}
int main(){
	while(~scanf("%d",&n)){
		if(n==0)break;
		for(int i=1;i<=n;i++){
			int x,y,z;
			scanf("%d%d%d",&villages[i].x,&villages[i].z,&villages[i].y);
		}
		init();
		double a,b;
		while(1){
			b=prim(a);
			if(fabs(b-a)<1e-4)break;
			a=b;
		}
		printf("%.3f
",a);
	}
	return 0;
}