GBX的Graph(最短路)

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Problem B: Graph

Time Limit: 2 Sec  Memory Limit: 128 MB
Submit: 1  Solved: 1
[Submit][Status][Web Board]

Description

There are n vertexs and m directed edges. Every vertex has a lowercase letters .Now, ZZT stand at 1.
he want to go to n to find ZZ.But ZZT dont like  character strings "cnm" ,"tmd","nsb". so  he won't walk such a continuous
3 vertex,the first vertex is 'c' second vertex is 'n'  last vertex is 'm'; He wanted to find her as soon as possible.

Input

The first line in the input will contain the number of cases ,Each case begins with two integer n,m(2<=n<=100,m<=1000)
then follow a line contain a character string "a1a2a3a4a5...an" ai is the i vertex's lowercase letter.
then follow m line ,each line contain li,ri,ci , a edge connect li and ri   cost time ci(1<=li,r1<=n,1<=ci<=100);

Output

for each case ,output a integer express the minimum time, if can't find ZZ output "-1"(without quotes);

Sample Input

1
4 4
cnmc
1 2 1
2 3 1
1 3 4
3 4 1

Sample Output

5

 

题意：给你一个n个点m条边的有向图，每一个点都有一个小写字母。现在ZZT站在点1，ZZ站在点n，ZZT想用最短的路程走到ZZ的地方。但是呢，ZZT不希望走过这样的连续的三点：cnm，tmd，nsb。现在问你他能否到达ZZ所在地。若能，输出最短路径，否则输出-1。

分析：此题关键在于如何构图。我们可以把边当点来使用，那么总共有m个点。加入一个源点0连向以点1发出去的边，加入一个汇点m+1使得点所有指向点n的边连向点m+1，那么原题就变成了从点0开始到达点m+1的最短路径。构图时，当两条边(u,v)，(x,y)中 v == x 并且(str[u],str[v],str[y]) !=(c,n,m),(t,m,d),(n,s,b)，则可连接。构图完毕，跑一遍最短路即可。

在其它博客无意中搜到的题，感觉挺巧妙的，把边看作了点，增加源点和汇点

原博客链接：https://blog.csdn.net/jhgkjhg_ugtdk77/article/details/47778893

原博客的题目链接失效了，找不提交的地方，就没写代码

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