思路:最短路
分析:只要把名字映射成整数,然后利用整数去求解即可。
注意事项:
1 题目中的起点和终点可能相同,这个时候输出0。
2 用map映射的时候用char类型,由于string是个类效率比较低。
3 处理成无向图
代码:
/*SPFA*/ #include<iostream> #include<algorithm> #include<cstdio> #include<string> #include<cstring> #include<queue> #include<map> using namespace std; #define MAXN 20010 #define INF 0XFFFFFFF int n , cnt; int begin , goal; int first[MAXN] , next[MAXN]; int star[MAXN] , end[MAXN] , value[MAXN]; int dis[MAXN]; queue<int>q; map<string , int>m; void init(){ cnt = 0; m.clear(); for(int i = 1 ; i <= 200 ; i++){ first[i] = -1; next[i] = -1; } } void SPFA(int s){ while(!q.empty()) q.pop(); int vis[MAXN]; memset(vis , 0 , sizeof(vis)); for(int i = 1 ; i <= cnt ; i++) dis[i] = INF; dis[s] = 0; q.push(s); vis[s] = 1; while(!q.empty()){ int x = q.front(); q.pop(); vis[x] = 0; for(int i = first[x] ; i != -1 ; i = next[i]){ if(dis[end[i]] > dis[x] + value[i]){ dis[end[i]] = dis[x] + value[i]; if(!vis[end[i]]){ vis[end[i]] = 1; q.push(end[i]); } } } } } void input(){ int a , b , v; char str1[200] , str2[200]; init(); scanf("%s%s" , str1 , str2); /*这里判断是不是相同点*/ a = m[str1]; if(!a) m[str1] = a = ++cnt; b = m[str2]; if(!b) m[str2] = b = ++cnt; begin = a; goal = b; for(int i = 0 ; i < n ; i++){ scanf("%s%s%d" , str1 , str2 , &v); a = m[str1]; if(!a) m[str1] = a = ++cnt; b = m[str2]; if(!b) m[str2] = b = ++cnt; /*处理成无向图*/ star[i] = a; end[i] = b; value[i] = v; star[i+n] = b; end[i+n] = a; value[i+n] = v; next[i] = first[star[i]]; first[star[i]] = i; next[i+n] = first[star[i+n]]; first[star[i+n]] = i+n; } } int main(){ while(scanf("%d" , &n) && n != -1){ input(); SPFA(begin); if(dis[goal] != INF) printf("%d\n" , dis[goal]); else printf("-1\n"); } return 0; }
/*floyd*/ #include<iostream> #include<algorithm> #include<cstdio> #include<string> #include<cstring> #include<queue> #include<map> using namespace std; #define MAXN 200 #define INF 0xFFFFFFF int n , star , end , cnt; long long dis[MAXN][MAXN]; map<string , int>m; long long min(long long a , long long b){ return a < b ? a : b; } void init(){ cnt = 0; m.clear(); for(int i = 1 ; i < MAXN ; i++){ for(int j = 1 ; j < MAXN ; j++){ if(i == j) dis[i][j] = 0; else dis[i][j] = INF; } } } void floyd(){ for(int k = 1 ; k <= cnt ; k++){ for(int i = 1 ; i <= cnt ; i++){ for(int j = 1 ; j <= cnt ; j++) dis[i][j] = min(dis[i][k]+dis[k][j] , dis[i][j]); } } } void input(){ int a , b , v; char str1[MAXN] , str2[MAXN]; init(); scanf("%s%s" , str1 , str2); a = m[str1]; if(!a) m[str1] = a = ++cnt; b = m[str2]; if(!b) m[str2] = b = ++cnt; star = a; end = b; for(int i = 0 ; i < n ; i++){ scanf("%s%s%d" , str1 , str2 , &v); a = m[str1]; b = m[str2]; if(!a) m[str1] = a = ++cnt; if(!b) m[str2] = b = ++cnt; if(dis[a][b] > v) dis[a][b] = dis[b][a] = v; } } int main(){ while(scanf("%d" , &n) && n != -1){ input(); floyd(); if(dis[star][end] != INF) printf("%d\n" , dis[star][end]); else printf("-1\n"); } return 0; }
时间: 2024-10-26 22:37:17