POJ 1502 MPI Maelstrom （简单最短路，多种算法均可）

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传送门：http://poj.org/problem?id=1502
题目大意：给定一个邻接矩阵，求出到结点1的最长的点的距离。
解题思路：简单题，n比较小，各种最短路算法均可，下面为SPFA。
Code：
/*   W          w           w        mm          mm             222222222       7777777777777    *//*    W        w w         w        m  m        m  m          222        22              7777    *//*    w        w w         w        m  m        m  m                     22              777     *//*     w      w   w       w        m    m      m    m                    22              77      *//*     w      w    w      w        m    m      m    m                 222                77      *//*      w    w      w    w        m      m    m      m              222                  77      *//*      w    w      w    w        m      m    m      m            222                    77      *//*       w  w        w  w        m        m  m        m         222                      77      *//*       w  w        w  w        m        m  m        m      222                         77      *//*        ww          ww        m          mm          m     222222222222222             77      *///#pragma comment(linker, "/STACK:102400000,102400000")//C++//int size = 256 << 20; // 256MB//char *p = (char*)malloc(size) + size;//__asm__("movl %0, %%esp\n" :: "r"(p));//G++#include<set>#include<map>#include<queue>#include<stack>#include<ctime>#include<deque>#include<cmath>#include<vector>#include<string>#include<cctype>#include<cstdio>#include<cstdlib>#include<cstring>#include<sstream>#include<iostream>#include<algorithm>#define REP(i,s,t) for(int i=(s);i<=(t);i++)#define REP2(i,t,s) for(int i=(t);i>=s;i--)using namespace std;typedef long long ll;typedef unsigned long long ull;typedef unsigned long ul;int n;const int MAXN=105;const int INF=0x3f3f3f3f;struct Edge{    int v;    int cost;    Edge(int _v=0,int _cost=0):v(_v),cost(_cost) {}};vector<Edge>E[MAXN];void addedge(int u,int v,int w){    E[u].push_back(Edge(v,w));}bool vis[MAXN];//在队列标志int cnt[MAXN];//每个点的入队列次数int dist[MAXN];bool SPFA(int start,int n){    memset(vis,false,sizeof(vis));    for(int i=1; i<=n; i++)dist[i]=INF;    vis[start]=true;    dist[start]=0;    queue<int>que;    while(!que.empty())que.pop();    que.push(start);    memset(cnt,0,sizeof(cnt));    cnt[start]=1;    while(!que.empty())    {        int u=que.front();        que.pop();        vis[u]=false;        for(int i=0; i<E[u].size(); i++)        {            int v=E[u][i].v;            if(dist[v]>dist[u]+E[u][i].cost)            {                dist[v]=dist[u]+E[u][i].cost;                if(!vis[v])                {                    vis[v]=true;                    que.push(v);                    if(++cnt[v]>n)return false;//cnt[i]为入队列次数,用来判定是否存在负环回路                }            }        }    }    return true;}char s[20];int main(){#ifdef ONLINE_JUDGE#else    freopen("test.in","r",stdin);#endif    while(~scanf("%d",&n))    {        REP(i,1,n)        {            E[i].clear();        }        REP(i,2,n)        {            REP(j,1,i-1)            {                scanf("%s",s);                if(s[0]=='x')                {                    continue;                }                else                {                    int w;                    sscanf(s,"%d",&w);                    addedge(i,j,w);                    addedge(j,i,w);                }            }        }        SPFA(1,n);        int ans=-1;        REP(i,2,n)        {            //printf("dist[%d]=%d\n",i,dist[i]);            if(dist[i]==INF)            {                continue;            }            ans=max(ans,dist[i]);        }        printf("%d\n",ans);    }    return 0;}
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