POJ 2240 && ZOJ 1082 Arbitrage 最短路,c++ stl pass g++ tle 难度:0

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

用log化乘法为加法找正圈

c++ 110ms,g++tle

#include <string>
#include <map>
#include <iostream>
#include <cmath>
#include <cstring>
#include <queue>
using namespace std;
const int maxn = 50;
bool vis[maxn];
double chg[maxn][maxn];
double dis[maxn];
int e[maxn][maxn],deg[maxn];
map<string,int> idmp;
int n,m;
const double inf = 0x3fffffff;
 
queue<int> que;
bool hasloop(int s){
    while(!que.empty())que.pop();
    que.push(s);
    vis[s] = true;
    int cnt = 0;
    while(!que.empty()){
        cnt ++;
        s = que.front();que.pop();vis[s] = false;
        for(int i = 0;i < deg[s];i++)
        {
            int t = e[s][i];
            if(dis[t] < dis[s] + chg[s][t])
            {
                dis[t] = dis[s] + chg[s][t];
                que.push(t);
                vis[t] = true;
            }
        }
        if(cnt > n * n)return true;
    }
    return false;
}
 
int main(){
    int ti = 0;
    while(cin>>n && n && ++ti){
        idmp.clear();
        for(int i = 0;i < n;i++)
        {
            dis[i] = -inf;
            for(int j = 0;j < n;j++)chg[i][j] = -inf;
        }
        memset(vis,false,sizeof vis);
        memset(deg,0,sizeof deg);
 
        for(int i = 0;i < n;i++)
        {
            string tmp;
            cin>>tmp;
            idmp[tmp] = i;
        }
        cin>>m;
        for(int i = 0;i < m;i++)
        {
            string sf,st;
            double change;
            cin>>sf>>change>>st;
            change = log(change);
            int f = idmp[sf];
            int t = idmp[st];
            chg[f][t] = change;
            e[f][deg[f]++] = t;
        }
        bool fl = false;
        for(int i = 0;i < n;i++)
        {
            if(dis[i] == -inf){
                dis[i] = 1;
                if(hasloop(i)){
                    fl = true;
                    break;
                }
            }
        }
        cout << "Case " << ti << ": ";
        if(fl)cout << "Yes" <<endl;
        else cout << "No" << endl;
    }
	return 0;
}