HDU4522 湫湫系列故事——过年回家

Posted 2020-11-03 Esquecer

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中文题面。

思路：拿spfa对卧铺和硬铺分别跑spfa，然后找两个的最短路。体感堆优化的dij也可以，不过spfa跑跑就过去了。有个细节是最后得用long long 存数据，其他的没啥。

           去重边是拿set存的邻接表。判断是否是数字用的isdigit函数。懒的要命系列。

#include <cstdio>  
#include <cstring>  
#include <cmath>  
#include <cstdlib>  
#include <ctime>  
#include <iostream>  
#include <algorithm>  
#include <sstream>  
#include <string>  
#include <vector>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <utility>  
#include <bitset>  

using namespace std;  
#define LL long long  
#define pb push_back  
#define mk make_pair  
#define pill pair<int, int>  
#define mst(a, b)    memset(a, b, sizeof a)  
#define REP(i, x, n)    for(int i = x; i <= n; ++i)
int vis[1000],dis[1000],n,m;
set<int>v1[210];
set<int>v2[210];
void spfa(int st,int key){
    queue<int>q;
    memset(vis,0,sizeof(vis));
    for(int i = 0 ; i <= n ; i++){
        dis[i] = 100000000;
    }
    q.push(st);
    vis[st] = 1;
    dis[st] = 0;
    while(!q.empty()){
        int v = q.front();
        vis[v] = 0 ;
        q.pop();
        if(key == 1){
            set<int>::iterator it = v1[v].begin();
            for(;it != v1[v].end(); it++){
                int to = *it;
                if(dis[to] > dis[v] + 1){
                    dis[to] = dis[v] + 1;
                    if(!vis[to]){
                        q.push(to);
                        vis[to] = 1;
                    }
                }
            }
        }
        else{
            set<int>::iterator it = v2[v].begin();
            for(;it != v2[v].end(); it++){
                int to = *it;
                if(dis[to] > dis[v] + 1){
                    dis[to] = dis[v] + 1;
                    if(!vis[to]){
                        q.push(to);
                        vis[to] = 1;
                    }
                }
            }
        }
    }
}
int main(){
    int t;
    for(scanf("%d",&t);t--;){
        char s[10010];
        scanf("%d %d",&n,&m);
        for(int i = 0 ; i <= n ; i++){
            v1[i].clear();
            v2[i].clear();    
        }
        while(m--){
            int num;
            scanf("%s %d",s,&num);
            int k = 0, y = -1;
            int len = strlen(s);
            for(int i = 0 ; i < len ; i++){
                if(isdigit(s[i])){
                    k = k * 10 + (s[i] - ‘0‘);
                }
                else{
                    if(y == -1) y = k;
                    else{
                        v1[y].insert(k);
                        if(num > 0){
                            v2[y].insert(k);
                        }
                        y = k;
                    }
                    k = 0;
                }
            }
            v1[y].insert(k);
            if(num > 0){
                v2[y].insert(k);
            }
        }
        long long ans1 = 0,ans2 = 0;
        int d1,d2,st,ed;
        scanf("%d %d %d %d",&d1,&d2,&st,&ed);
        spfa(st,1);
        ans1 = dis[ed];
        spfa(st,0);
        ans2 = dis[ed];
        if(ans1 == 100000000 && ans2 == 100000000 ){
            puts("-1");continue;
        }
        //printf("%d %d\n",ans1,ans2);
        printf("%lld\n",min(ans1*d1*1LL,ans2*d2*1LL));
    }
}