Poj_2240 Arbitrage(最短路)

来源：互联网发布：苹果网络锁破解编辑：程序博客网时间：2024/05/17 22:31

题意：

套利就是利用汇率的差异使自己的财富得到增加。给出几种货币之间的汇率，问能否通过套利使得财富增加。

思路：

转换为寻找有向图中是否有环。使用Bellman-Ford和SPFA分别实现了一下，SPFA稍微快一点，注意SPFA是入队超过N次。

代码实现：

//Bellman-Ford#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <map>#include <algorithm>using namespace std;const int MAX_V = 32;struct Edge{    int from;    int to;    double rate;};int N,M;int cnt;bool flag;double curr[MAX_V];map<string,int> mm;Edge edge[MAX_V*MAX_V];bool find_loop();int main(){    cnt = 1;    while( scanf("%d",&N) != EOF ){        if( N == 0 ){            break;        }        string a,b;        mm.clear();        for( int i = 0; i < N; i++ ){            cin>>a;            mm[a] = i+1;        }        double rate;        scanf("%d",&M);        for( int i = 0; i < M; i++ ){            cin>>a>>rate>>b;            edge[i].from = mm[a];            edge[i].to = mm[b];            edge[i].rate = rate;        }        flag = find_loop();        if( flag == true ){            printf("Case %d: Yes\n",cnt++);        }        else{            printf("Case %d: No\n",cnt++);        }    }    return 0;}bool find_loop(){    memset(curr,0,sizeof(curr));    curr[1] = 1;    for( int i = 0; i < N; i++ ){        for( int j = 0; j < M; j++ ){            Edge e = edge[j];            if( curr[e.to] < curr[e.from]*e.rate ){                curr[e.to] = curr[e.from]*e.rate;                if( i == N-1 ){                    return true;                }            }        }    }    return false;}//SPFA#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <map>#include <vector>#include <queue>using namespace std;const int MAX_V = 32;struct Edge{    int to;    double rate;    Edge(const int to,const double rate){        this->to = to;        this->rate = rate;    }};int N,M;int cnt;bool flag;int num[MAX_V];queue<int> que;bool inque[MAX_V];map<string,int> mm;double curr[MAX_V];vector<Edge> edge[MAX_V];int main(){    cnt = 1;    while( scanf("%d",&N) != EOF ){        if( N == 0 ){            break;        }        string a,b;        mm.clear();        for( int i = 0; i < MAX_V; i++ ){            edge[i].clear();        }        for( int i = 0; i < N; i++ ){            cin>>a;            mm[a] = i+1;        }        double rate;        scanf("%d",&M);        for( int i = 0; i < M; i++ ){            cin>>a>>rate>>b;            int pa = mm[a];            int pb = mm[b];            edge[pa].push_back(Edge(pb,rate));        }        flag = false;        memset(num,0,sizeof(num));        memset(curr,0,sizeof(curr));        memset(inque,false,sizeof(inque));        curr[1] = 1;        que.push(1);        inque[1] = true;        num[1]++;        while( !que.empty() ){            int tmp = que.front();            que.pop();            inque[tmp] = false;            if( flag == true ){                break;            }            int len = edge[tmp].size();            for( int i = 0; i < len; i++ ){                Edge e = edge[tmp][i];                if( curr[e.to] < curr[tmp]*e.rate ){                    curr[e.to] = curr[tmp]*e.rate;                    if( inque[e.to] == false ){                        inque[e.to] = true;                        num[e.to]++;                        que.push(e.to);                        if( num[e.to] > N ){                            flag = true;                            break;                        }                    }                }            }        }        if( flag == true ){            printf("Case %d: Yes\n",cnt++);        }        else{            printf("Case %d: No\n",cnt++);        }    }    return 0;}

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