POJ-2240 arbitrage Bellman

Basically the same as POJ-1860, bellman asks whether there is a ring. Here, we learned other people's code to add the External Loop of the bellman algorithm directly, and then determine whether there is any update or release in this process.

The Code is as follows:

#include <cstring>#include <cstdlib>#include <cstdio>#include <map>#include <string>using namespace std;int N, pos, cnt;double dis[35];map<string,int>mp;struct Node{    int x, y;    double rate;}e[905];bool bellman(){    int flag;    fill(dis, dis+35, 1000);    for (int i = 1; i <= N; ++i) {        flag = 0;        for (int j = 1; j <= pos; ++j) {            if (dis[ e[j].x ] * e[j].rate - dis[ e[j].y ] > 1e-6) {                dis[ e[j].y ] = dis[ e[j].x ] * e[j].rate;                flag = 1;            }        }        if (!flag) {            break;        }    }    if (flag) {        return true;    }    else {        return false;    }}int main(){    int M, ca = 0;    char a[105], b[105];    double rate;    while (scanf("%d", &N), N) {        mp.clear();        pos = cnt = 0;        for (int i = 1; i <= N; ++i) {            scanf("%s", a);            mp[a] = ++cnt;        }        scanf("%d", &M);        for (int i = 1; i <= M; ++i) {            scanf("%s %lf %s", a, &rate, b);            ++pos;            e[pos].x = mp[a], e[pos].y = mp[b];            e[pos].rate = rate;        }        printf("Case %d: ", ++ca);        printf(bellman() ? "Yes\n" : "No\n");    }    return 0;}