poj1502 spfa最短路

  1 //Accepted    320 KB    16 ms
  2 //有n个顶点,边权用A表示
  3 //给出下三角矩阵，求从一号顶点出发到各点的最短路的最大值
  4 #include <cstdio>
  5 #include <cstring>
  6 #include <iostream>
  7 #include <queue>
  8 #include <cmath>
  9 #include <algorithm>
 10 using namespace std;
 11 /**
 12   * This is a documentation comment block
 13   * 如果有一天你坚持不下去了，就想想你为什么走到这儿！
 14   * @authr songt
 15   */
 16 const int imax_n = 105;
 17 const int imax_e = imax_n*imax_n;
 18 const int inf = 100000000;
 19 struct node
 20 {
 21     int u,v,c;
 22     node(int u=0,int v=0,int c=0):u(u),v(v),c(c)
 23     {
 24 
 25     }
 26 }p[imax_e];
 27 int head[imax_n];
 28 int next[imax_e];
 29 bool vis[imax_n];
 30 int dis[imax_n];
 31 int cnt[imax_n];
 32 int n;
 33 int e=0;
 34 void init()
 35 {
 36     memset(head,-1,sizeof(head));
 37     memset(next,-1,sizeof(next));
 38     e=0;
 39 }
 40 void addEdge(int u,int v,int c)
 41 {
 42     p[e]=node(u,v,c);
 43     next[e]=head[u];
 44     head[u]=e++;
 45 }
 46 bool relax(int u,int v,int c)
 47 {
 48     if (dis[v]>dis[u]+c)
 49     {
 50         dis[v]=dis[u]+c;
 51         return true;
 52     }
 53     return false;
 54 }
 55 queue<int > q;
 56 bool spfa(int src)
 57 {
 58     while (!q.empty()) q.pop();
 59     memset(vis,0,sizeof(vis));
 60     memset(cnt,0,sizeof(cnt));
 61     for (int i=1;i<=n;i++)
 62     dis[i]=inf;
 63     dis[src]=0;
 64     q.push(src);
 65     vis[src]=1;
 66     cnt[src]++;
 67     while (!q.empty())
 68     {
 69         int pre=q.front();
 70         q.pop();
 71         vis[pre]=0;
 72         for (int i=head[pre];i+1;i=next[i])
 73         {
 74             if (relax(pre,p[i].v,p[i].c) && !vis[p[i].v])
 75             {
 76                 if ((++cnt[p[i].v])>=n) return false;
 77                 vis[p[i].v]=1;
 78                 q.push(p[i].v);
 79             }
 80         }
 81     }
 82     return true;
 83 }
 84 char s[10];
 85 int main()
 86 {
 87     while (scanf("%d",&n)!=EOF)
 88     {
 89         init();
 90         for (int i=2;i<=n;i++)
 91         {
 92             for (int j=1;j<i;j++)
 93             {
 94                 scanf("%s",s);
 95                 if (s[0]==‘x‘) continue;
 96                 int c=atoi(s);
 97                 //printf("c=%d\n",c);
 98                 addEdge(i,j,c);
 99                 addEdge(j,i,c);
100             }
101         }
102         spfa(1);
103         int ans=0;
104         for (int i=1;i<=n;i++)
105         if (dis[i]>ans) ans=dis[i];
106         printf("%d\n",ans);
107     }
108     return 0;
109 }