[2016-04-02][POJ][2387][Til the Cows Come Home]

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• 时间:2016-04-02 10:34:36 星期六

• 题目编号:[2016-04-02][POJ][2387][Til the Cows Come Home]

• 题目大意:给定n个节点和t条路,求n到1的最短路长度

• 分析:跑一次最短路即可

• 遇到的问题:

  • 据说是多重边,如果是用邻接矩阵的就要更新最小值,
  • 此题是先输入t,再输入n,输入的时候读错,无限WA…

#include <queue>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 1000 + 10;
const int maxe = 2000 + 10;
struct mNode{
        int u,c;
        mNode(int _u = 0,int _c = 0):u(_u),c(_c){}
        bool operator < (const mNode & a)const{
                return c > a.c;
        }
};
struct Edge{
        int v,c;
        Edge(int _v = 0,int _c = 0):v(_v),c(_c){}
};
vector<Edge> e[maxe];
bool vis[maxn];
int d[maxn];
void Dijkstra(int s){
        memset(vis,0,sizeof(vis));
        memset(d,0x3f,sizeof(d));
        priority_queue<mNode> q;
        d[s] = 0;
        q.push(mNode(s,0));
        mNode tmp;
        while(!q.empty()){
                tmp = q.top();
                q.pop();
                int u = tmp.u;
                if(vis[u])      continue;
                vis[u] = 1;
                for(int i = 0;i < e[u].size();++i){
                        Edge & me = e[u][i];
                        if(!vis[me.v] && d[me.v] > d[u] + me.c){
                                d[me.v] = d[u] + me.c;
                                q.push(mNode(me.v,d[me.v]));
                        }                      
                }
        }
}
inline void addedge(int u,int v,int c){
        e[u].push_back(Edge(v,c));
}
int main(){
        int n,t,a,b,c;
        scanf("%d%d",&t,&n);
        for(int i = 0;i < t; ++i){
                scanf("%d%d%d",&a,&b,&c);
                addedge(a,b,c);
                addedge(b,a,c);
        }
        Dijkstra(n);
        printf("%d\n",d[1]);
        return 0;
}

[2016-04-02][POJ][2387][Til the Cows Come Home]

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原文地址：http://www.cnblogs.com/qhy285571052/p/fa56a3d106dbda764c3e4164a5177926.html