基础最短路 九 POJ 2240

POJ 2240：http://poj.org/problem?id=2240

n为 30

依然是前面的类似的货币交换问题，问能否靠这种交换货币的方法发家致富.......

预处理部分用map映射成数字，接下来就好办了。

Spfa：

#include"cstdio"#include"iostream"#include"cstring"#include"algorithm"#include"vector"#include"cmath"#include"queue"#include"map"using namespace std;#define INF 9999999#define inf 1009#define loop(x,y,z) for(x=y;x<z;x++)#define ll long longint book[inf];int cnt[inf];int n,m;double dis[inf];struct node{    int to;    double rate;    node(int i,double j)    {        to=i;        rate=j;    }};vector<node>edge[inf];queue<int>q;map<string,int>currency;void setmap(){    int i;    string t;    loop(i,0,n)    {        cin>>t;        currency[t]=i;    }}void init(){    int i;    memset(book,0,sizeof book);    memset(cnt,0,sizeof cnt);    loop(i,0,n)dis[i]=-1;    dis[0]=100;    loop(i,0,n)edge[i].clear();    currency.clear();    while(!q.empty())q.pop();}bool spfa(){    q.push(0);    book[0]=1;    int i;    while(!q.empty())    {        int u=q.front();        q.pop();        book[u]=0;        int len=edge[u].size();        loop(i,0,len)        {            node& e=edge[u][i];            if(dis[e.to]<dis[u]*e.rate)            {                dis[e.to]=dis[u]*e.rate;                if(!book[e.to])                {                    q.push(e.to);                    book[e.to]=1;                    if(++cnt[e.to]>n-1)return false;                }            }        }    }    return true;}int main(){    int i=1;    string a,b;    int x,y;    double c,t;    while(~scanf("%d",&n)&&n)    {        init();        setmap();        scanf("%d",&m);        while(m--)        {            cin>>a>>c>>b;            x=currency[a];            y=currency[b];            edge[x].push_back(node(y,c));        }        if(spfa())cout<<"Case "<<i++<<": No"<<endl;        else cout<<"Case "<<i++<<": Yes"<<endl;    }    return 0;}