Til the Cows Come Home POJ

题目：https://vjudge.net/problem/POJ-2387

好久没写博客了，算自己懒吧。

最短路入门题，自己练习用SPFA实现了下，不需要判负环也写上了。WA了好几次，忘记判断重边了。

代码：C++

//SPFA#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#include <set>#include <map>using namespace std;const int maxn = 2000 + 10;const int INF = 2147483647;map<int, map<int, int> > pic;int T, n;queue<int> q;bool inq[maxn];int cnt[maxn];int dis[maxn];void init(){    memset(cnt, 0, sizeof(cnt));    memset(inq, false, sizeof(inq));    for(int i = 1; i <= n; i++)    {        dis[i] = INF;    }    dis[1] = 0;}bool SPFA(){    q.push(1);    inq[1] = true;    cnt[1]++;    while(!q.empty())    {        int cur = q.front();        q.pop();        inq[cur] = false;        for(map<int, int>::iterator iter = pic[cur].begin(); iter!=pic[cur].end(); iter++)        {            if(dis[cur]!=INF && dis[cur] + iter->second < dis[iter->first])            {                dis[iter->first] = dis[cur] + iter->second;                if(!inq[iter->first])                {                    q.push(iter->first);                    cnt[iter->first]++;                    inq[iter->first] = true;                    //判断负环                    if(cnt[iter->first] > n)                    {                        return false;                    }                }            }        }    }    return true;}int main(){    scanf("%d%d", &T, &n);    init();    while(T--)    {        int x, y, w;        scanf("%d%d%d", &x, &y, &w);        if(!pic[x].count(y))        {            pic[x][y] = w;            pic[y][x] = w;        }        else        {            pic[x][y] = min(w, pic[x][y]);            pic[y][x] = min(w, pic[y][x]);        }    }    SPFA();    printf("%d\n", dis[n]);    return 0;}