最短路径（两个权值且具有优先级)

http://acm.hdu.edu.cn/showproblem.php?pid=3790

题意：给出两个权值的图，距离和价值，距离优先于价值。求u、v间最短路径。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll ;
#define int ll
#define mod 998244353
#define gcd(m,n) __gcd(m, n)
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
int lcm(int a , int b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define size(v) (int)(v.size())
#define cin(x) scanf("%lld" , &x);
const int N = 1e4+9;
const int maxn = 1e3+9;
const double esp = 1e-6;
int ma[maxn][maxn] , vis[maxn] , dis[maxn] , val[maxn];
int mi[maxn][maxn] ;
int n , m ;


void dijkstra(int u){
    rep(i , 1 , n){
        dis[i] = ma[u][i];
        val[i] = mi[u][i];
    }
    vis[u] = 1;
    rep(i , 1 , n-1){
        int pos ;
        int mix = INF;
        int miv = INF;
        rep(j , 1 , n){
            if(!vis[j] && mix > dis[j]){
                mix = dis[j];
                miv = val[j];
                pos = j ;
            }else if(!vis[j] && mix == dis[j] && miv > val[j]){
                miv = val[j];
                pos = j ;
            }
        }
        vis[pos] = 1 ;
        rep(j , 1 , n){
            if(!vis[j] && dis[j] > dis[pos] + ma[pos][j]){
                dis[j] = dis[pos] + ma[pos][j];
                val[j] = val[pos] + mi[pos][j];
            }else if(!vis[j] && dis[j] == dis[pos] + ma[pos][j]){
                val[j] = min(val[j] , val[pos] + mi[pos][j]);
            }
        }
    }
}
void init(){
    fill(ma[0] , ma[0] + maxn*maxn , INF);
    fill(mi[0] , mi[0] + maxn*maxn , INF);
    ME(vis , 0);
}

void solve(){
    init();
    rep(i , 1 , m){
        int u , v , w , va;
        scanf("%lld%lld%lld%lld" , &u , &v , &w , &va);
        if(ma[u][v] > w){//这里要注意可能出现重边，可能出现距离小的价值大，距离大的价值小，直接都取最小的话，没有对应起来出错。
            ma[u][v] = ma[v][u] = w ;
            mi[u][v] = mi[v][u] = va;
        }else if(ma[u][v] == w){
            mi[u][v] = va;
        }
    }
    int u , v ;
    scanf("%lld%lld" , &u , &v);
    dijkstra(u);
    cout << dis[v] << " " << val[v] << endl;
}

signed main()
{
    while(~scanf("%lld%lld" , &n , &m) && n+m)
        solve();
}

 堆优化：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll ;
#define int ll
#define mod 998244353
#define gcd(m,n) __gcd(m, n)
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
int lcm(int a , int b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define size(v) (int)(v.size())
#define cin(x) scanf("%lld" , &x);
const int N = 1e5+9;
const int maxn = 1e3+9;
const double esp = 1e-6;
int head[maxn] , tol ;
int dis[maxn] , vis[maxn] , val[maxn];
int n , m ;
struct Graph{
    int v , w , p , next;
}g[N<<1];
void add(int u , int v , int w , int p){
    g[++tol] = {v , w , p , head[u]};
    head[u] = tol;
}
struct Edge{
    int v , w , p;
    bool operator < (const Edge &e) const{
        if(w > e.w){
            return true;
        }else if(w < e.w){
            return false;
        }else{
            return p > e.p;
        }
    }
    Edge(int _v , int _w , int _p){
        v = _v , w = _w , p = _p;
    }
    Edge(){}
};

void dijkstra(int u){
    fill(dis , dis+maxn , INF);
    fill(val , val+maxn , INF);
    dis[u] = 0 ; val[u] = 0 ;
    priority_queue<Edge>q;
    q.push(Edge(u , dis[u] , val[u]));
    Edge now;
    while(!q.empty()){
        now = q.top() ; q.pop();
        vis[now.v] = 1 ;
        for(int i = head[now.v] ; i ; i = g[i].next){
            int v = g[i].v , w = g[i].w , p = g[i].p;
            if(!vis[v] && dis[v] > dis[now.v] + w){
                dis[v] = dis[now.v] + w ;
                val[v] = val[now.v] + p ;
                q.push(Edge(v , dis[v] , val[v]));
            }else if(!vis[v] && dis[v] == dis[now.v] + w && val[v] > val[now.v] + p){
                val[v] = val[now.v] + p;
                q.push(Edge(v , dis[v] , val[v]));
            }
        }
    }
}
void init(){
    ME(vis , 0);
    ME(head , 0);
    tol = 0 ;
}

void solve(){
    init();
    rep(i , 1 , m){
        int u , v , w , p ;
        scanf("%lld%lld%lld%lld" , &u , &v , &w , &p);
        add(u , v , w , p);
        add(v , u , w , p);
    }
    int u , v ;
    scanf("%lld%lld" , &u , &v);
    dijkstra(u);
    cout << dis[v] << " " << val[v] << endl;
}

signed main()
{
    while(~scanf("%lld%lld" , &n , &m) && n+m)
        solve();
}