• poj2728 Desert King


    懒得学 Dinkelbach迭代法……二分卡过

    完全图就不要用 (mathrm{O}(mlog m)) 的堆优化了……

    #include <algorithm>
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <cmath>
    using namespace std;
    int n, uu[1005], vv[1005], ww[1005], din;
    double a[1005][1005], dis[1005], qaq[1005][1005], pap[1005][1005];
    const double eps=1e-8;
    bool vis[1005];
    struct Node{
    	int idx;
    	double val;
    }nd[1100005];
    double qwq(int u, int v){
    	return sqrt((uu[u]-uu[v])*(uu[u]-uu[v])+(vv[u]-vv[v])*(vv[u]-vv[v]));
    }
    bool cmp(Node x, Node y){
    	return x.val>y.val;
    }
    bool chk(double lim){
    	for(int i=1; i<=n; i++)
    		for(int j=1; j<=n; j++)
    			a[i][j] = pap[i][j] - lim * qaq[i][j];
    	memset(vis, 0, sizeof(vis));
    	memset(dis, 127, sizeof(dis));
    	dis[1] = din = 0;
    	nd[++din] = (Node){1, 0.0};
    	double re=0.0;/*
    	while(din){
    		Node x=nd[1];
    		pop_heap(nd+1, nd+1+din, cmp);
    		din--;
    		if(vis[x.idx])	continue;
    		vis[x.idx] = true;
    		re += dis[x.idx];
    		for(int i=1; i<=n; i++)
    			if(!vis[i] && dis[i]>a[x.idx][i]){
    				dis[i] = a[x.idx][i];
    				nd[++din] = (Node){i, dis[i]};
    				push_heap(nd+1, nd+1+din, cmp);
    			}
    	}*/
    	for(int ii=1; ii<=n; ii++){
    		int maxi=0;
    		double orz=100000000.0;
    		for(int i=1; i<=n; i++)
    			if(!vis[i] && dis[i]<orz){
    				orz = dis[i];
    				maxi = i;
    			}
    		vis[maxi] = true;
    		re += orz;
    		for(int i=1; i<=n; i++){
    			if(!vis[i] && dis[i]>a[maxi][i])
    				dis[i] = a[maxi][i];
    		}
    	}
    	return re<=eps;
    }
    int main(){
    	while(scanf("%d", &n)!=EOF){
    		if(!n)	break;
    		for(int i=1; i<=n; i++)
    			scanf("%d %d %d", &uu[i], &vv[i], &ww[i]);
    		for(int i=1; i<=n; i++)
    			for(int j=1; j<=n; j++){
    				qaq[i][j] = qwq(i, j);
    				pap[i][j] = abs(ww[i]-ww[j]);
    			}
    		double l=0.0, r=10000000.0, mid, re;
    		while(fabs(r-l)>eps){
    			mid = (l + r) / 2.0;
    			if(chk(mid))	re = mid, r = mid;
    			else	l = mid;
    		}
    		printf("%.3f
    ", re);
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/poorpool/p/8607607.html
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