HDU 2112 迪杰斯特拉最短路算法

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题目 HDU 2112
迪杰斯特拉最短路 实际上就是优先队列的广搜，每次提取队列中距离最小的进行搜索

AC代码如下，用了 map set 处理字符串很方便

#include<iostream>
#include<string>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<cstdio>
using namespace std;
const int maxn = 166;
int d[maxn];
vector< pair<int,int> > v[maxn];
void init(){
    for(int i = 0;i < maxn;++i){
        v[i].clear();
        d[i] = 1 << 30;
    }
}
int main(){
    int n;
    while(scanf("%d",&n) != EOF){
        if(n == -1) break;
        init();
        set<string> S;
        map<string,int> Map;
        string str,str0;int ssize = 0;
        cin >> str;S.insert(str);if(S.size() > ssize) Map[str] = ssize ++;
        cin >> str0;S.insert(str0);if(S.size() > ssize) Map[str0] = ssize ++;
        string str1,str2;int t;
        for(int i = 0;i < n;++i){
            cin >> str1 >> str2 >> t;
            S.insert(str1);if(S.size() > ssize) Map[str1] = ssize ++;
            S.insert(str2);if(S.size() > ssize) Map[str2] = ssize ++;
            int a = Map[str1];
            int b = Map[str2];
            v[a].push_back(make_pair(b,t));
            v[b].push_back(make_pair(a,t));
        }
        priority_queue< pair<int,int> > Q;
        int st = Map[str];
        d[st] = 0;Q.push(make_pair(-d[st],st));
        while(!Q.empty()){
            int now = Q.top().second;Q.pop();
            for(int i = 0;i < v[now].size();++i){
                int u = v[now][i].first;
                if(d[u] > d[now]+v[now][i].second){
                    d[u] = d[now]+v[now][i].second;
                    Q.push(make_pair(-d[u],u));
                }
            }
        }
        int ed = Map[str0];
        if(d[ed] == 1 << 30) printf("%d\n",-1);
        else printf("%d\n",d[ed]);
    }
    return 0;
}