SPFA找最大比例环

SPFA找最大比例环#

ans=Σ点权/Σ边权

所以 可以变式为 Σ边权*ans-Σ点权=0

要找出最大的ans 可以二分

边权值变为 目的地点权-ans*边权

检查是否有负环 有则可以更优

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define CLR(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int maxn = 5010;
const double EPS = 1e-6;
int n,m,val[maxn],from[maxn],to[maxn],w[maxn];
int head[maxn],tol;
int vis[maxn],c[maxn],d[maxn];
double dis[maxn];

struct edge
{
    int from,to,next;
    double w;
}edge[maxn];

void addedge(int u,int v,double w)
{
    edge[tol].from = u;
    edge[tol].to = v;
    edge[tol].w = w;
    edge[tol].next = head[u];
    head[u] = tol++;
}

void init()
{
    CLR(head,-1);
    CLR(vis,0);
    CLR(c,0);
    CLR(d,0);
    for(int i = 0; i <= n; i++) dis[i] = 1000001;
    tol = 0;
}

bool check(double ans)
{
    init();
    for(int i = 1; i <= m; i++){
        int u = from[i];
        int v = to[i];
        addedge(u,v,(1.0*val[v] - 1.0*w[i]*ans)*-1);
        d[u]++;
    }
    for(int i = 1; i <= n; i++){
        if(d[i] == 0) addedge(0,i,0);
    }
    addedge(0,1,0);

    queue<int> q;
    q.push(0);
    dis[0] = 0;
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = 0;

        for(int i = head[u]; ~i; i = edge[i].next){
            int v = edge[i].to;
            double W = edge[i].w;
            if(dis[u] + W < dis[v]){
                dis[v] = W + dis[u];
                if(!vis[v]){
                    vis[v] = 1;
                    c[v]++;
                    if(c[v] > n){
                        return false;
                    }
                    q.push(v);
                }
            }
        }
    }

    return true;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n; i++) scanf("%d",&val[i]);
    for(int i = 1; i <= m; i++){
        scanf("%d%d%d",&from[i],&to[i],&w[i]);
    }

    double l = 0;
    double r = 1000001;

    while( r - l > EPS){
        double mid = (l + r) / 2;
        if(check(mid))
            r = mid;
        else
            l = mid;
    }

    printf("%.2f\n",r);

    return 0;
}