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  • https://vjudge.net/problem/Gym-101630J/origin

    先跑一遍spfa,这样就排除了小于k条边的情况。枚举那k条边中最小的边权,然后让所有的边都减去它,接着跑spfa,然后在加上k*v[i]

    #include <iostream>
    #include <cstdio>
    #include <queue>
    #include <algorithm>
    #include <cmath>
    #include <cstring>
    #define inf 2147483647000000
    #define N 1000010
    #define p(a) putchar(a)
    #define For(i,a,b) for(register long long i=a;i<=b;++i)
    
    using namespace std;
    long long n,m,k,x,y,ans;
    long long v[N],d[N];
    bool vis[N];
    deque<long long>q;
    struct node{
        long long n;
        long long v;
        node *next;
    }*e[N];
    
    inline void in(long long &x){
        long long y=1;char c=getchar();x=0;
        while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
        while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
        x*=y;
    }
    inline void o(long long x){
        if(x<0){p('-');x=-x;}
        if(x>9)o(x/10);
        p(x%10+'0');
    }
    
    inline void push(register long long x,register long long y,register long long v){
        node *p;
        p=new node();
        p->n=y;
        p->v=v;
        if(e[x]==0)
            e[x]=p;
        else{
            p->next=e[x]->next;
            e[x]->next=p;
        }
    }
    
    inline long long spfa(long long t){
        For(i,0,n) vis[i]=0,d[i]=inf;
        d[1]=0;
        q.push_back(1);
        while(!q.empty()){
            x=q.front();q.pop_front();
            vis[x]=1;
            for(register node *i=e[x];i;i=i->next){
                if(d[i->n]>d[x]+max((long long)0,i->v-t)){
                    d[i->n]=d[x]+max((long long)0,i->v-t);
                    if(!vis[i->n]){
                        vis[i->n]=1;
                        if(!q.empty()&&d[i->n]<d[q.front()]) q.push_front(i->n);
                        else q.push_back(i->n);
                    }    
                }
            }
    
            vis[x]=0;
        }
        return d[n];
    }
    
    signed main(){
        in(n);in(m);in(k);
        For(i,1,m){
            in(x);in(y);in(v[i]);
            push(x,y,v[i]);
            push(y,x,v[i]);
        }
        ans=spfa(0);
        For(i,1,m) ans=min(ans,spfa(v[i])+k*v[i]);
        o(ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/war1111/p/12570604.html
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