spfa+01 规划

尼玛的哪里错了。。

/*
在有向图上找一个环，使结点权值和/边权和的比例值最大
01规划，设比例为l，那么将每条边的权值改成a[u]-l*w，如果有正权环，则比例l可行
如何判图中存在正权环？将 权值存相反数，spfa判负环即可 
复杂度elogans 
*/
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
#define maxn 1005
#define maxm 5005
#define esp 0.0000001
#define INF 0x3f3f3f3f
struct Edge{int from,to,nxt,w;}edge[maxm],e[maxm];
int head[maxn],tot,a[maxn],n,m;
void init(){
    memset(head,-1,sizeof head);
    tot=0;
}
void addedge(int u,int v,int w){
    edge[tot].from=u,e[tot].from=u;
    edge[tot].to=v,e[tot].to=v;
    edge[tot].nxt=head[u],e[tot].nxt=head[u];
    edge[tot].w=w,head[u]=tot++;
}

double d[maxn];
int v[maxn],cnt[maxn];
int judge(double mid){
    for(int i=0;i<tot;i++){
        int u=edge[i].from;
        e[i].w=-(1.0*a[u]-mid*edge[i].w);
    }
    //spfa
    for(int i=1;i<=n;i++)d[i]=INF*1.0;
    memset(v,0,sizeof v);
    memset(cnt,0,sizeof cnt);
    d[1]=0;v[1]=1;
    queue<int>q;
    q.push(1);cnt[1]=1;
    while(!q.empty()){
        int x=q.front();q.pop();
        v[x]=0;
        for(int i=head[x];i!=-1;i=edge[i].nxt){
            int y=edge[i].to,z=e[i].w;
            if(d[y]>d[x]+z){
                d[y]=d[x]+z;
                if(v[y]==0)q.push(y),v[y]=1;
                if(++cnt[y]>n)return 1;
            }
        }
    }
    return 0;
}

int main(){
    while(scanf("%d%d",&n,&m)==2){
        init();
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        for(int i=1;i<=m;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
        }
        double l=0.0,r=100.0,mid;
        while(l+esp<=r){
            mid=(l+r)/2;
            if(judge(mid))l=mid;
            else r=mid;
        }
        printf("%.2lf
",l);
    }    
    return 0;
}

下面这样的就没问题了。。

#include<cstdio>
#include<cstring>
#include<queue>
#define eps 1e-3
#define MAXN 1010
using namespace std;
struct T
{
    int v;
    int w;
    int next;
}edge[5005];
int cnt,head[MAXN];
void add_edge(int u,int v,int w)
{
    edge[cnt].v = v;
    edge[cnt].w = w;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}
int n,m;
int w[MAXN];
double dis[MAXN];
bool inque[MAXN];
int vis[MAXN];
bool SPFA(double R)//SPFA判断负权环
{
    queue<int> myque;
    memset(inque,0,sizeof inque);
    memset(vis,0,sizeof vis);
    for(int i = 1; i <= n; i++)
        dis[i] = 1e15;
    myque.push(1);
    dis[1] = 0;
    inque[1] = 1;
    vis[1]++;
    while(!myque.empty())
    {
        int u = myque.front();
        myque.pop();
        inque[u] = 0;
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].v;
            int y = edge[i].w;
            if(dis[u] + R*y - w[u] < dis[v])
            {
                dis[v] = dis[u] + R*y - w[u];
                if(!inque[v])
                {
                    myque.push(v);
                    inque[v] = 1;
                    vis[v]++;
                    if(vis[v] > n) return 1;
                }
            }
        }
    }
    return 0;
}
int main()
{
    memset(head,-1,sizeof head);
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n; i++)
        scanf("%d",&w[i]);
    for(int i = 1; i <= m; i++)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        add_edge(a,b,c);
    }
    double l = 0,r = 10000,mid;
    while(r - l > eps)
    {
        mid = (l+r)/2;
        if(SPFA(mid)) l = mid;//由于我们是把权值取反了的，因此题解中的R过大变成了R过小
        else r = mid;
    }
    printf("%.2lf
",l);
    return 0;
}