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;}

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