POJ-1502 MPI Maelstrom

题目链接：https://vjudge.net/problem/POJ-1502

给出n个节点的连通图，求从1点到其他所有点的最短路，再输出其中最远的那个，这里用dijkstra做

输入以邻接矩阵下三角的方式给出，注意如果没有相连以字符‘x’表示，处理时需要一些输入技巧

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<sstream>using namespace std;const int N=100+2;const int inf=0x7f7f7f7f;struct node{int pos;int dist;node(int a,int b){pos=a;dist=b;}bool operator < (const node &u) const{return dist<u.dist;}};int n;int g[N][N],d[N];int main(){string a;int b;while(~scanf("%d",&n)){memset(g,0,sizeof(g));for(int i=0;i<n;i++)for(int j=0;j<i;j++){cin>>a;if(a[0]=='x')g[i][j]=g[j][i]=inf;else{stringstream s;s<<a;s>>b;g[i][j]=g[j][i]=b;}}memset(d,inf,sizeof(d));priority_queue<node> q;q.push(node(0,0));d[0]=0;while(!q.empty()){node u=q.top();q.pop();int p=u.pos;for(int i=0;i<n;i++)if(d[i]>d[p]+g[p][i]){d[i]=d[p]+g[p][i];q.push(node(i,d[i]));}}int ans=0;for(int i=1;i<n;i++)if(d[i]>ans) ans=d[i];printf("%d\n",ans);}return 0;}