01 分数规划

P2868 Sightseeing Cows G

给出一个 \(n\) 个点 \(m\) 条的有点权与边权的有向图，求一个环使得环上的点权和除以边权和最大。

最优比率环问题,显然01分数规划

环上第\(i\)个点点权为\(a_i\),第\(i\)条边的边权为\(b_i\)

\[\frac{\sum_{i=1}^nF_i}{\sum_{i=1}^mT_i}=ans\\ 对于01分数规划,考虑二分.二分可以将最优化问题转化为判定问题.如果二分出来mid为L,则问题转化为是否存在\\ \frac{\sum F_i}{\sum T_i}>L~~~分母乘过去(T_i>0)\\ \sum(F_i-L*T_i)>0~~左式乘-1~~~\sum(L*T_i-F_i)<0\\ 问题转化为求负环,对于入点为u_i,出点为v_i的边e_i，把边权看做mid*T[e_i]-F[u_i],判负环\\ 有负环则L=mid,否则R=mid \]

#include<cstdio>
#include<queue>
#include<iostream>
#define maxm 5005
#define maxn 1002
using namespace std;
inline int read() {
    int x = 0,f = 1; char s = getchar();
    while(s < '0' || s > '9') {if(s == '-') f = -1;s = getchar();}
    while(s >= '0' && s <= '9') {x = x * 10 + s - '0';s = getchar();}
    return f * x;
}
int l,p,head[maxn],tot = 0,vis[maxn],num[maxn],a[maxn];
double d[maxn];
struct edge{
	int to,next,dis;
}e[maxm];
inline void add(int u,int v,int w){
	e[++tot].to = v;
	e[tot].next = head[u];
	e[tot].dis = w;
	head[u] = tot;
}
int check(double x){//找负环 
	queue<int> q;
	for(int i = 1;i <= l;i++){
		q.push(i);
		d[i] = 0;vis[i] = num[i] = 1;
	}
	while(!q.empty()){
		int u = q.front();
		q.pop();vis[u] = 0;
		for(int i = head[u];i;i = e[i].next){
			int v = e[i].to;
			double dis = (double)e[i].dis;
			if(d[v] > d[u] + x * dis - (double)a[u])//边权为mid*Tim[e_i]-Fun[u_i]
            {
                d[v] = d[u] + x * dis - (double)a[u];
                if(!vis[v])
                {
                    q.push(v); vis[v]=1;
                    if(++num[v] >= l) return 1;
                }
            }
		} 
	}
	return 0;
}
int main(){
 l=read();p=read();
    for(int i=1;i<=l;++i)a[i]=read();
    for(int i=1;i<=p;++i)
    {
        int u=read(),v=read(),dis=read();
        add(u,v,dis);
    }
    double L=0,R=5001,mid;
    while(R-L>1e-4)
    {
        mid=(L+R)/2;
        if(check(mid))L=mid;
        else R=mid;
    }
    printf("%.2lf",L);
    return 0;
}

P4377 [USACO18OPEN]Talent Show G

乍眼一看,这题好裸,裸题就意味着神题(裸体就意味着身体)

\[1.分数规划\\ ans=\frac{\sum t_i}{\sum w_i}(\sum w_i\ge W)\\ \frac{\sum t_i}{\sum w_i}\ge mid \\ 所以要求\sum (t_i-mid*w_i)\ge0,令c_i=t_i-mid*w_i\\ 2.~~01背包,做个大小为W的背包,重量大于dp[w]的,都记在dp[w]上,最后只要看dp[W]是否>0 \]

#include<cstdio>
#include<iostream>
#define INF 0x3f3f3f3f
#define maxn 251
#define maxm 1005
using namespace std;
int n, W;
int w[maxn],t[maxn];
double dp[maxm],c[maxn];
inline int max(int a,int b){return a > b ? a : b;}
inline int check(double mid){
	for(int i = 1;i <= n;i++) c[i] = (double)t[i] - mid * w[i];
	for(int i = 1;i <= W;i++)	dp[i] = -INF;
	for(int i = 1;i <= n;i++)
		for(int j = W;j >= 0;j--){
			if(j + w[i] >= W) dp[W]=max(dp[W], dp[j] + c[i]);
			else dp[j+w[i]]=max(dp[j+w[i]],dp[j]+c[i]);
		}
	return dp[W]>=0;
}
int main(){
	scanf("%d%d",&n,&W);
	double l=0,r=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&w[i],&t[i]);
		r+=t[i];
	}
	while(l+1e-5<=r)
	{
		double mid=(l+r)/2;
		if(check(mid))
		{
			l=mid;	
		}
		else 
			r=mid;
	}
	int ans = (int)(l*1000);
	printf("%d",ans);
	return 0;
}