題意
題解
01分數規劃,點權爲
01分數規劃講解
用了二分法。
實現時用到了spfa的dfs版本判負環,精度1E-4,交G++ 0MS。
代碼
/// by ztx
/// blog.csdn.net/hzoi_ztx
#include <bits/stdc++.h>
#define Rep(i,l,r) for(i=(l);i<=(r);i++)
#define rep(i,l,r) for(i=(l);i< (r);i++)
#define Rev(i,r,l) for(i=(r);i>=(l);i--)
#define rev(i,r,l) for(i=(r);i> (l);i--)
#define Each(i,v) for(i=v.begin();i!=v.end();i++)
#define r(x) read(x)
typedef long long ll ;
typedef double lf ;
int CH , NEG ;
template <typename TP>inline void read(TP& ret) {
ret = NEG = 0 ; while (CH=getchar() , CH<'!') ;
if (CH == '-') NEG = true , CH = getchar() ;
while (ret = ret*10+CH-'0' , CH=getchar() , CH>'!') ;
if (NEG) ret = -ret ;
}
#define kN 1010LL
#define kM 5010LL
#define t(p) e[0][p]
#define n(p) e[1][p]
#define c(p) e[2][p]
#define eps 1E-4
#define infi 11111111LL
int e[3][kM<<1], st[kN] = {0}, te = 1;
inline void AddEdge(int u,int v,int w) {
te ++ , t(te) = v, c(te) = w, n(te) = st[u], st[u] = te; // 單向
// te ++ , t(te) = u, c(te) = w, n(te) = st[v], st[v] = te;
}
int n, v[kN];
lf L, M, R, dis[kN];
bool vis[kN];
bool SPFA(int u) { // return true when it has a nag-ring
vis[u] = true;
for (int p = st[u]; p; p = n(p))
if (dis[t(p)] > dis[u]+M*c(p)-v[u])
if (dis[t(p)] = dis[u]+M*c(p)-v[u], vis[t(p)]) return true;
else if (SPFA(t(p))) return true;
vis[u] = false;
return false;
}
// spfa bfs判負環 需要n個點都進隊,否則如果不連通就gg了
// spfa dfs則從每個點dfs
inline bool Neg() {
int i;
memset(dis,0,sizeof dis);
memset(vis,0,sizeof vis);
Rep (i,1,n) if (SPFA(i)) return true;
return false;
}
int main() {
int m, i, u, v, w;
r(n), r(m);
Rep (i,1,n) r(::v[i]);
Rep (i,1,m)
r(u), r(v), r(w), AddEdge(u,v,w);
L = 0, R = 1000.0;
while (R-L > eps) { // L < r* , R >= r*
M = (L+R)/2;
if (Neg()) L = M;// have neg-rings
else R = M;
}
printf("%.2f\n", R);
END: getchar(), getchar();
return 0;
}