Uva11374 Airport Express

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最短路问题。

从起点和终点开始各跑一次dijkstra,可以得到起点、终点到任意点的距离。枚举使用的商业线路,找最优解。

破题卡输出,记录前驱和输出什么的仿佛比算法本身还麻烦。

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;
const int mxn=2410;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<0 || ch>9){if(ch==-)f=-1;ch=getchar();}
    while(ch>=0 && ch<=9){x=x*10+ch-0;ch=getchar();}
    return x*f;
}
int n,m,k,S,E;
struct exp{
    int x,y,w;
}ex[mxn];
struct edge{
    int v,nxt,dis;
}e[mxn<<2];
int hd[mxn],mct=0;
void add_edge(int u,int v,int d){
    e[++mct].v=v;e[mct].nxt=hd[u];e[mct].dis=d;hd[u]=mct;
    return;
}
struct node{
    int v,d;
    bool operator < (const node r) const {return d>r.d;}
};
priority_queue<node>tp;
int pre[mxn],nxt[mxn];
int dis[mxn];
int mdis[mxn];
void Dij1(int st){
    while(!tp.empty())tp.pop();
    memset(dis,0x3f,sizeof dis);
    tp.push((node){st,0});
    dis[st]=0;
    while(!tp.empty()){
        node tmp=tp.top();tp.pop();
        if(tmp.d>dis[tmp.v])continue;
        int u=tmp.v;
        for(int i=hd[u];i;i=e[i].nxt){
            int v=e[i].v;
            if(dis[v]>dis[u]+e[i].dis){
                dis[v]=dis[u]+e[i].dis;
                pre[v]=u;
                tp.push((node){v,dis[v]});
            }
        }
    }
    return;
}
void Dij2(int st){
    while(!tp.empty())tp.pop();
    memset(dis,0x3f,sizeof dis);
    tp.push((node){st,0});
    dis[st]=0;
    while(!tp.empty()){
        node tmp=tp.top();tp.pop();
        if(tmp.d>dis[tmp.v])continue;
        int u=tmp.v;
        for(int i=hd[u];i;i=e[i].nxt){
            int v=e[i].v;
            if(dis[v]>dis[u]+e[i].dis){
                dis[v]=dis[u]+e[i].dis;
                nxt[v]=u;
                tp.push((node){v,dis[v]});
            }
        }
    }
    return;    
}
int st[mxn],top=0;
void init(){
    memset(hd,0,sizeof hd);
    memset(pre,0,sizeof pre);
    memset(nxt,0,sizeof nxt);
    mct=0;top=0;
}

int main(){
    int i,j;
    bool flag=0;
    while(scanf("%d%d%d%d",&n,&S,&E,&m)!=EOF){
        if(flag){
            printf("\n");
        }
        flag=1;
        init();
        int u,v,w;
        for(i=1;i<=m;i++){
            u=read();v=read();w=read();
            add_edge(u,v,w);
            add_edge(v,u,w);
        }
        k=read();
        for(i=1;i<=k;i++){
            ex[i].x=read();ex[i].y=read();
            ex[i].w=read();
        }
        Dij1(S);
        memcpy(mdis,dis,sizeof dis);
        Dij2(E);
        int pos1=0,pos2=0,ans=1e9;
        for(i=1;i<=n;i++){
            if(ans>mdis[i]+dis[i]){
                ans=mdis[i]+dis[i];
                pos1=i;
            }
        }
        for(i=1;i<=k;i++){
            int tmp=mdis[ex[i].x]+dis[ex[i].y]+ex[i].w;
            if(ans>tmp){
                ans=tmp;pos1=ex[i].x;pos2=ex[i].y;
            }
            tmp=mdis[ex[i].y]+dis[ex[i].x]+ex[i].w;
            if(ans>tmp){
                ans=tmp;pos1=ex[i].y;pos2=ex[i].x;
            }
        }
        if(pos1 && !pos2){
//            printf("Test1\n");
            int tmp=pos1;
            while(pre[pos1]){
                pos1=pre[pos1];
                st[++top]=pos1;
            }
            while(top)    printf("%d ",st[top--]);
            pos1=tmp;
            do{
                printf("%d",pos1);
                if(nxt[pos1])printf(" ");
                pos1=nxt[pos1];
            }while(pos1);
            printf("\nTicket Not Used\n");
        }
        else{
            if(pos1 && pos2){
                int tmp=pos1;
                while(pre[pos1]){
                    pos1=pre[pos1];
                    st[++top]=pos1;
                }
                while(top)    printf("%d ",st[top--]);
                pos1=tmp;
                printf("%d ",pos1);
//                pos2=nxt[pos2];
                while(pos2){
                    printf("%d",pos2);
                    if(nxt[pos2])printf(" ");
                    pos2=nxt[pos2];
                }
                printf("\n%d\n",pos1);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

 

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