首页 > 代码库 > poj3268 Silver Cow Party (SPFA求最短路)

poj3268 Silver Cow Party (SPFA求最短路)

其实还是从一个x点出发到所有点的最短路问题。来和回只需分别处理一下逆图和原图,两次SPFA就行了。

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cmath>#include<map>#include<set>#include<vector>#include<algorithm>#include<stack>#include<queue>#include<cctype>#include<sstream>using namespace std;#define pii pair<int,int>#define LL long long intconst int eps=1e-8;const int INF=1000000000;const int maxn=1000+10;int n,m,x,a,b,t;vector<int>vv1[maxn];vector<int>vv2[maxn];int t1[maxn][maxn],t2[maxn][maxn];int d1[maxn],d2[maxn],ans=0;//以上1是原图的量,2是逆图的量queue<int>q;void spfa(int type){    if(type==1)    {        q.push(x);        while(!q.empty())        {            int tt=q.front();            q.pop();            int st=vv1[tt].size();            for(int i=0;i<st;i++)            {                int ti=vv1[tt][i];                if(d1[tt]+t1[tt][ti]<d1[ti])                {                    d1[ti]=d1[tt]+t1[tt][ti];                    q.push(ti);                }            }        }    }    else    {        q.push(x);        while(!q.empty())        {            int tt=q.front();            q.pop();            int st=vv2[tt].size();            for(int i=0;i<st;i++)            {                int ti=vv2[tt][i];                if(d2[tt]+t2[tt][ti]<d2[ti])                {                    d2[ti]=d2[tt]+t2[tt][ti];                    q.push(ti);                }            }        }    }}int main(){    //freopen("in8.txt","r",stdin);    //freopen("out.txt","w",stdout);    scanf("%d%d%d",&n,&m,&x);    fill(d1,d1+maxn,INF);    fill(d2,d2+maxn,INF);    d1[x]=d2[x]=0;    for(int i=0;i<m;i++)    {        scanf("%d%d%d",&a,&b,&t);        t1[a][b]=t2[b][a]=t;        vv1[a].push_back(b);        vv2[b].push_back(a);    }    spfa(1);    spfa(2);    for(int i=1;i<=n;i++)    {        ans=max(ans,d1[i]+d2[i]);    }    printf("%d\n",ans);    //fclose(stdin);    //fclose(stdout);    return 0;}

 

poj3268 Silver Cow Party (SPFA求最短路)