哈理工练习赛 UVALive 5093 F - Seaside（最短路）

F - Seaside

Time Limit:3000MS    Memory Limit:0KB    64bit IO Format:%lld & %llu

SubmitStatusPracticeUVALive 5093

Description

XiaoY is living in a big city, there are N towns in it and some towns near the sea. All these towns are numbered from 0 toN - 1 and XiaoY lives in the town numbered '0'. There are some directed roads connecting them. It is guaranteed that you can reach any town from the town numbered '0', but not all towns connect to each other by roads directly, and there is no ring in this city. One day, XiaoY want to go to the seaside, he asks you to help him find out the shortest way.

Input

There are several test cases. In each cases the first line contains an integerN(0N10), indicating the number of the towns. Then followed N blocks of data, in block-i there are two integers,M_i(0M_iN - 1) and P_i, then M_i lines followed. M_i means there areM_i roads beginning with the i-th town. P_i indicates whether thei-th town is near to the sea, P_i = 0 means `No', P<sub>i</sub> = 1 means `Yes'. In nextM_i lines, each line contains two integersS_Mi and L_Mi, which means that the distance between thei-th town and the S_Mi town isL_Mi.

Output

Each case takes one line, print the shortest length that XiaoY reach seaside.

Sample Input

5 1 0 1 1 2 0 2 3 3 1 1 1 4 100 0 1 0 1

Sample Output

2题意： 给出一个无向图，多出的仅仅是某个结点是否临海，让我们判断从起点0开始，到最近的临海点距离。思路：完全可以在输入的时候记录所有临海的点，dij一遍后，在这些点中找出最小距离值即可。AC ：
/*=============================================================================##      Author: liangshu - cbam ##      QQ : 756029571 ##      School : 哈尔滨理工大学 ##      Last modified: 2015-09-06 14:39##     Filename: hdu1015.cpp##     Description: #        The people who are crazy enough to think they can change the world, are the ones who do ! =============================================================================*/#    #include<iostream>    #include<sstream>    #include<algorithm>    #include<cstdio>    #include<string.h>    #include<cctype>    #include<string>    #include<cmath>    #include<vector>    #include<stack>    #include<queue>    #include<map>    #include<set>    using namespace std;    const int maxn=30003;    const int INF=0x1f1f1f1f;    int n,m;    //标记起始位置    string s;    string e;     set<int >xx;    struct Edge    {        int from,to,dist;        Edge(int u,int v,int d):from(u),to(v),dist(d) {}    };    vector<Edge>edges;//存储边的结点信息    vector<int>G[maxn];//邻接表    bool done[maxn];   //标记是否访问过    int d[maxn];    //距离数组    struct heapnode //用于优先队列自定义    {        int d,u;        bool operator <(const heapnode rhs) const        {            return d > rhs.d;        }        heapnode(int dd,int uu):d(dd),u(uu) {}    };    void init()//初始化必不可少    {        for(int i=0; i<n; i++)            G[i].clear();        edges.clear();    }    void addedge(int from,int to,int dist)    {        edges.push_back(Edge(from,to,dist));        int m = edges.size();        G[from].push_back(m-1);    }    void dij( int s)    {        priority_queue<heapnode>Q;        for(int i=0; i<=n; i++)//初始化距离数组            d[i]=INF;        d[s]=0;        memset(done ,0,sizeof(done));        Q.push( heapnode(0,s) );        while(!Q.empty())        {            heapnode x = Q.top();            Q.pop();            int u=x.u;            if(done[u])continue;            done[u] = true;            for(int i=0; i<G[u].size(); i++)            {                Edge& e=edges[G[u][i]];//取出来一条邻接边                //cout<<"u = "<<u<<" e.u = "<<e.to<<endl;                if(d[e.to]>d[u]+e.dist)                {                    d[e.to] = d[u] + e.dist;                    Q.push((heapnode(d[e.to],e.to)));                }            }        }    }    int main()    {        while(scanf("%d",&n)!=EOF)        {            int t=1;            int ans=0;            init();           xx.clear();int a, b;            for(int i=0; i<n  ; i++)            {                scanf("%d%d",&a,&b);                if(b == 1){                    xx.insert(i);                }                if(a == 0)continue;                for(int z = 0; z< a; z++){                    int x, y;                    scanf("%d%d",&x, &y);                     addedge(i, x,y);                addedge(x,i ,y);                }            }            dij(0);            int M = 0x1f1f1f1f;            for(int i = 0; i < n  ;i++){                    if(xx.count(i))              {                    M = min(M, d[i]);              }            }            cout<<M<<endl;            s.clear();            e.clear();        }        return 0;    }/*51 01 12 02 33 51 14 1000 10 1    */