P4951 [USACO01OPEN]Earthquake(二分&MST)
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P4951 [USACO01OPEN]Earthquake(二分&MST)
01分数规划。
f − ∑ c o s t i ∑ t i ≥ a n s \\dfracf-\\sum cost_i\\sum t_i\\ge ans ∑tif−∑costi≥ans
f − ( a n s × ∑ t i + ∑ c o s t i ) ≥ 0 f-(ans\\times \\sum t_i+\\sum cost_i)\\ge 0 f−(ans×∑ti+∑costi)≥0
边权改为 − ( a n s × t i + c o s t i ) -(ans\\times t_i+cost_i) −(ans×ti+costi)
二分答案 a n s ans ans。
求最大生成树,如果大于等于0,说明答案可以更大。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=505,M=1e4+5,inf=0x3f3f3f3f,mod=1e9+7;
const int hashmod[4] = 402653189,805306457,1610612741,998244353;
#define mst(a,b) memset(a,b,sizeof a)
#define db double
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define x first
#define y second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define ios ios::sync_with_stdio(false),cin.tie(nullptr)
void Print(int *a,int n)
for(int i=1;i<n;i++)
printf("%d ",a[i]);
printf("%d\\n",a[n]);
template <typename T> //x=max(x,y) x=min(x,y)
void cmx(T &x,T y)
if(x<y) x=y;
template <typename T>
void cmn(T &x,T y)
if(x>y) x=y;
struct Kruskal
#define il inline
struct edge
int u,v,t;
db w;
bool operator<(const edge &e)const
return w>e.w;
e[M],e1[M];
int s[N];
const db eps = 1e-6;
il int find(int x)
return s[x]==x?x:s[x]=find(s[x]);
int n,m;
db f;
il void init()
scanf("%d%d%lf",&n,&m,&f);
rep(i,1,n) s[i] = i;
rep(i,1,m)
scanf("%d%d%lf%d",&e[i].u,&e[i].v,&e[i].w,&e[i].t);
il void solve()
double l = 0, r = f;
while(r-l>eps)
db mid = (l+r)/2;
rep(i,1,m) e1[i] = e[i] , e1[i].w = -e[i].w - e[i].t*mid;
rep(i,1,n) s[i] = i;
if(ck()) l = mid;
else r = mid;
printf("%.4f\\n",l);
il bool ck()
sort(e1+1,e1+m+1);
int cnt = 0;
db ans = 0;
rep(i,1,m)
int u = e1[i].u, v = e1[i].v;
db w = e1[i].w;
u=find(u),v=find(v);
if(u!=v)
ans += w;
s[u] = v;
cnt++;
if(cnt == n - 1)
break;
return f+ans>=0;
T;
int main()
T.init();
T.solve();
return 0;
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