GBX的Graph(最短路)

Problem B: Graph

Time Limit: 2 Sec Memory Limit:
128 MB

Submit: 1 Solved: 1

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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

Sample Output

题意：给你一个n个点m条边的有向图，每个点都有一个小写字母。

如今ZZT站在点1。ZZ站在点n。ZZT想用最短的路程走到ZZ的地方。可是呢，ZZT不希望走过这种连续的三点：cnm，tmd，nsb。如今问你他是否能到达ZZ所在地。

若能，输出最短路径，否则输出-1。

分析：此题关键在于怎样构图。

我们能够把边当点来使用，那么总共同拥有m个点。

增加一个源点0连向以点1发出去的边，增加一个汇点m+1使得点全部指向点n的边连向点m+1，那么原题就变成了从点0開始到达点m+1的最短路径。构图时，当两条边(u,v)。(x,y)中 v == x 而且(str[u],str[v],str[y]) !=(c,n,m),(t,m,d),(n,s,b)，则可连接。

构图完成。跑一遍最短路就可以。

题目链接：http://192.168.4.140/problem.php?

cid=1000&pid=1

代码清单：

#include<set>

#include<map>

#include<cmath>

#include<queue>

#include<stack>

#include<ctime>

#include<cctype>

#include<string>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

typedef long long ll;

typedef unsigned int uint;

typedef unsigned long long ull;

const int maxn = 100 + 5;

const int maxv = 1000 + 5;

const int max_dis=0x7f7f7f7f;

struct Edge{

    int to;

    int dis;

    Edge(){}

    Edge(int to,int dis){

        this -> to = to;

        this -> dis = dis;

    }

};

struct E{ int u,v,dis; }edge[maxv];

int T;

int n,m;

int a,b,c;

bool vis[maxv];

int dist[maxv];

char str[maxn];

vector<Edge>graph[maxv];

void init(){

    for(int i=0;i<maxv;i++) graph[i].clear();

}

void input(){

    scanf("%d%d%s",&n,&m,str);

    for(int i=1;i<=m;i++){

        scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].dis);

    }

}

bool judge(int pre,int now,int to){

    if(str[pre]=='c'&&str[now]=='n'&&str[to]=='m') return true;

    if(str[pre]=='t'&&str[now]=='m'&&str[to]=='d') return true;

    if(str[pre]=='n'&&str[now]=='s'&&str[to]=='b') return true;

    return false;

}

void createGraph(){

    for(int i=1;i<=m;i++){

        if(edge[i].u==1) graph[0].push_back(Edge(i,edge[i].dis));

        if(edge[i].v==n) graph[i].push_back(Edge(m+1,0));

        for(int j=1;j<=m;j++){

            if(i==j) continue;

            if(edge[i].v==edge[j].u){

                if(judge(edge[i].u-1,edge[i].v-1,edge[j].v-1))

                    continue;

                graph[i].push_back(Edge(j,edge[j].dis));

            }

        }

    }

}

int spfa(){

    memset(vis,false,sizeof(vis));

    memset(dist,max_dis,sizeof(dist));

    queue<int>q;

    while(!q.empty()) q.pop();

    vis[0]=true; dist[0]=0;

    q.push(0);

    while(!q.empty()){

        int u=q.front(); q.pop();

        vis[u]=false;

        for(int i=0;i<graph[u].size();i++){

            int v=graph[u][i].to;

            int d=graph[u][i].dis;

            if(dist[v]>dist[u]+d){

                dist[v]=dist[u]+d;

                if(!vis[v]){

                    vis[v]=true;

                    q.push(v);

                }

            }

        }

    }

    if(dist[m+1]==max_dis) return -1;

    return dist[m+1];

}

void solve(){

    createGraph();

    printf("%d\n",spfa());

}

int main(){

//    freopen("cin.txt","r",stdin);

//    freopen("cout.txt","w",stdout);

    scanf("%d",&T);

    while(T--){

        init();

        input();

        solve();

    }return 0;

}