Luogu4926 倍杀测量者（二分答案+差分约束）

　　容易想到二分答案。问题变为判断是否所有条件都被满足，可以发现这是很多变量间的相对关系，取个log之后就是经典的差分约束模型了。特殊的地方在于某些人的分数已被给定，从每个人开始跑一遍最短路判断一下是否能满足关系即可。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 1010
const double eps=1E-6;
double l,r,ans,d[N],a[N];
int n,m,k,p[N],q[N],cnt[N],t;
bool f[N],isget[N];
struct data{int to,nxt;double len;
}edge[N<<2];
struct flag{int op,x,y,z;
}Q[N];
void addedge(int x,int y,double z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;}
int inc(int &x){x++;if (x>n+1) x-=n+1;return x;}
bool spfa(int k)
{
    memset(f,0,sizeof(f));
    for (int i=1;i<=n;i++) d[i]=10000000;d[k]=0;
    memset(cnt,0,sizeof(cnt));
    int head=0,tail=1;q[1]=k;
    do
    {
        int x=q[inc(head)];f[x]=0;
        for (int i=p[x];i;i=edge[i].nxt)
        if (d[x]+edge[i].len<d[edge[i].to])
        {
            d[edge[i].to]=d[x]+edge[i].len;
            if (!f[edge[i].to])
            {
                q[inc(tail)]=edge[i].to,f[edge[i].to]=1;
                cnt[edge[i].to]++;if (cnt[edge[i].to]==n) return 0;
            }
        }
    }while (head!=tail);
    if (isget[k])
    for (int i=1;i<=n;i++)
    if (isget[i]&&d[i]<a[i]-a[k]) return 0;
    return 1;
}
bool check(double T) 
{
    t=0;memset(p,0,sizeof(p));
    for (int i=1;i<=m;i++)
    if (Q[i].op==1)
    {
        if (Q[i].z>T) addedge(Q[i].x,Q[i].y,-log(Q[i].z-T));
    }
    else addedge(Q[i].x,Q[i].y,log(Q[i].z+T));
    for (int i=1;i<=n;i++)
    if (!spfa(i)) return 1;
    return 0;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("c.in","r",stdin);
    freopen("c.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read(),k=read();
    for (int i=1;i<=m;i++)
    Q[i].op=read(),Q[i].x=read(),Q[i].y=read(),Q[i].z=read();
    l=eps,r=10-eps;ans=-1;
    for (int i=1;i<=k;i++)
    {
        int x=read(),y=read();
        a[x]=log(y);isget[x]=1;
    }
    while (l<=r)
    {
        double mid=(l+r)/2;
        if (check(mid)) ans=mid,l=mid+eps;
        else r=mid-eps;
    }
    if (ans<0) cout<<-1;
    else printf("%.8lf",ans);
    return 0;
}

 

转载于:https://www.cnblogs.com/Gloid/p/9762749.html