HDU - 1546 ZOJ - 2750 Idiomatic Phrases Game 成语接龙SPFA+map
Idiomatic Phrases Game
InputThe input consists of several test cases. Each test case contains an idiom dictionary. The dictionary is started by an integer N (0 < N < 1000) in one line. The following is N lines. Each line contains an integer T (the time Tom will take to work out) and an idiom. One idiom consists of several Chinese characters (at least 3) and one Chinese character consists of four hex digit (i.e., 0 to 9 and A to F). Note that the first and last idioms in the dictionary are the source and target idioms in the game. The input ends up with a case that N = 0. Do not process this case.
OutputOne line for each case. Output an integer indicating the shortest time Tome will take. If the list can not be built, please output -1.Sample Input
5 5 12345978ABCD2341 5 23415608ACBD3412 7 34125678AEFD4123 15 23415673ACC34123 4 41235673FBCD2156 2 20 12345678ABCD 30 DCBF5432167D 0
Sample Output
17 -1
题意:每个成语由至少三个字组成,每个字为四个16进制的数字(0~F)组成。两个成语首尾相接,消耗Tom对前一个成语的查询时间,求最小查询时间。
思路:最短路问题,注意对建图的处理。map将成语的首尾字以数字点的形式存储。使用map前必须初始化。
#include<stdio.h> #include<string.h> #include<string> #include<map> #include<deque> #include<vector> #define MAX 2005 #define INF 0x3f3f3f3f using namespace std; map<string,int> mp; struct Node{ int v,w; }node; vector<Node> edge[MAX]; int dis[MAX],b[MAX],w[MAX],bg[MAX],ed[MAX]; int n; void spfa(int k) { int i; deque<int> q; for(i=1;i<=n;i++){ dis[i]=INF; } memset(b,0,sizeof(b)); b[k]=1; dis[k]=0; q.push_back(k); while(q.size()){ int u=q.front(); for(i=0;i<edge[u].size();i++){ int v=edge[u][i].v; int w=edge[u][i].w; if(dis[v]>dis[u]+w){ dis[v]=dis[u]+w; if(b[v]==0){ b[v]=1; if(dis[v]>dis[u]) q.push_back(v); else q.push_front(v); } } } b[u]=0; q.pop_front(); } } int main() { int m,u,v,i,j; char s[200],s1[200],s2[200]; while(scanf("%d",&n)&&n!=0){ m=0; for(i=1;i<=2002;i++){ edge[i].clear(); mp.clear(); //!! } for(i=1;i<=n;i++){ scanf("%d%s",&w[i],s); strncpy(s1,s,4); strncpy(s2,s+strlen(s)-4,4); s1[4]='\0';s2[4]='\0'; if(!mp[s1]) mp[s1]=++m; if(!mp[s2]) mp[s2]=++m; u=mp[s1];v=mp[s2]; bg[i]=u;ed[i]=v; } for(i=1;i<=n;i++){ for(j=1;j<=n;j++){ if(i!=j&&ed[i]==bg[j]){ node.v=j; node.w=w[i]; edge[i].push_back(node); } } } spfa(1); if(n==1||dis[n]==INF) printf("-1\n"); else printf("%d\n",dis[n]); } return 0; }
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· .NET 9 new features-C#13新的锁类型和语义
· Linux系统下SQL Server数据库镜像配置全流程详解
· 现代计算机视觉入门之:什么是视频
· 你所不知道的 C/C++ 宏知识
· 聊一聊 操作系统蓝屏 c0000102 的故障分析
· 不到万不得已,千万不要去外包
· C# WebAPI 插件热插拔(持续更新中)
· 会议真的有必要吗?我们产品开发9年了,但从来没开过会
· 如何打造一个高并发系统?
· 【译】我们最喜欢的2024年的 Visual Studio 新功能