GBX的Graph(最短路)

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。

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;
}