POJ2240 Arbitrage 货币汇率

/*
*题目大意：
*        有n种货币，而且每两种之间有汇率，如果a->b,b->c,c->a,然后他们之间的
*        所有汇率乘积大于1，那么就是一种获得利润的手段，题目要求输出yes.
*解题思路：
*        其实画图就知道，题目要的就是找到存在一个圈，并且这个圈所有权值的乘积
*        是正的，如果把最长路的关系条件由加法改为乘法，那么就是说找一个所有乘
*        积为的环。dij,floyd可以办到吗？显然不行，他们都处理不了带环的。这个时候
*        spfa跟Bellman_ford就可以发挥无可比拟的优势了，不过囧了，这道题居然spfa
*        比优化了的Bellman_ford慢了那么多，Bellman_ford 47ms,spfa要219ms
*解题感想：
*        记住想想咯，spfa的话，一个点入队n次，就马上判断有不可收敛的环了，入队n
*        次，代表了人家松弛了n-1次。
*/

Bellman_ford:

#include <iostream>
#include <queue>
#include <map>
#include <string>
#include <cstdio>
#include <cstring>
using namespace std;

const int MAXN = 35;
const int MAXE = 100005;
const double inf = 0x3f3f3f3f;

typedef struct _node
{
    int u, v;
    double w;
    int next;
}N;
map<string, int> index;
N edge[MAXE];
int head[MAXN], cntEdge;

void init(int n)
{
    index.clear();
    cntEdge = 0;
    for(int i = 0; i <= n; i++)
    {
        head[i] = -1;
    }
}

void addEdge(int u, int v, double w)
{
    edge[cntEdge].u = u;
    edge[cntEdge].v = v;
    edge[cntEdge].w = w;
    edge[cntEdge].next = head[u];
    head[u] = cntEdge++;
}

bool Bellman_ford(int s, int n)
{
    double dis[MAXN];
    for(int i = 0; i < n; i++)
    {
        dis[i] = -inf;
    }
    dis[s] = 1;

    for(int i = 0; i < n-1; i++)
    {
        bool flag = false;
        for(int j = 0; j < cntEdge; j++)
        {
            int u = edge[j].u, v = edge[j].v;
            double w = edge[j].w;
            if(dis[v] < dis[u] * w)
            {
                dis[v] = dis[u] * w;
                flag = true;
            }
        }
        if(!flag)
            break;
    }

    for(int f = 0; f < cntEdge; f++)
    {
        int u = edge[f].u, v = edge[f].v;
        double w = edge[f].w;
        if(dis[v] < dis[u] * w)
            return true;//有可以增值的环
    }
    return false;
}

int main(void)
{
#ifndef ONLINE_JUDGE
    //freopen("in.txt", "r", stdin);
#endif

    int n, m, cas_c = 1;
    while(scanf("%d", &n), n)
    {
        init(n);
        int cnt = 0;
        char name[1024], name1[1024];
        for(int i = 0; i < n; i++)
        {    
            scanf("%s", name);
            if(!index.count(name))
                index[name] = cnt++;
        }
        
        scanf("%d", &m);
        double w;
        for(int i = 0; i < m; i++)
        {
            scanf("%s %lf %s", &name, &w, &name1);
            addEdge(index[name], index[name1], w);
        }

        //正图
        bool sol = Bellman_ford(0, n);
        if(sol)
            printf("Case %d: Yes\n", cas_c++);
        else
            printf("Case %d: No\n", cas_c++);
    }
    return 0;
}

spfa:

#include <iostream>
#include <queue>
#include <map>
#include <queue>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;

const int MAXN = 35;
const int MAXE = 100005;
const double inf = 0x3f3f3f3f;

typedef struct _node
{
    int u, v;
    double w;
    int next;
}N;
map<string, int> index;
N edge[MAXE];
int head[MAXN], cntEdge;

void init(int n)
{
    index.clear();
    cntEdge = 0;
    for(int i = 0; i <= n; i++)
    {
        head[i] = -1;
    }
}

void addEdge(int u, int v, double w)
{
    edge[cntEdge].u = u;
    edge[cntEdge].v = v;
    edge[cntEdge].w = w;
    edge[cntEdge].next = head[u];
    head[u] = cntEdge++;
}

bool spfa(int s, int n)
{
    int inQ[MAXN] = {0}, in[MAXN] = {0};
    double dis[MAXN];
    for(int i = 0; i < n; i++)
        dis[i] = -inf;
    
    dis[s] = 1;
    queue<int> Q;
    Q.push(s);
    inQ[s] = 1;
    in[s]++;

    while(!Q.empty())
    {
        int pre = Q.front();
        Q.pop();
        inQ[pre] = 0;

        for(int f = head[pre]; f != -1; f = edge[f].next)
        {
            int v = edge[f].v;
            double w = edge[f].w;
            if(dis[v] < dis[pre] * w)
            {
                dis[v] = dis[pre] * w;
                if(!inQ[v])
                {
                    inQ[v] = 1;
                    Q.push(v);
                    in[v]++;
                    if(in[v] > n)
                        return true;
                }
            }
        }
    }
    return false;
}

int main(void)
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif

    int n, m, cas_c = 1;
    while(scanf("%d", &n), n)
    {
        init(n);
        int cnt = 0;
        char name[1024], name1[1024];
        for(int i = 0; i < n; i++)
        {    
            scanf("%s", name);
            if(!index.count(name))
                index[name] = cnt++;
        }
        
        scanf("%d", &m);
        double w;
        for(int i = 0; i < m; i++)
        {
            scanf("%s %lf %s", &name, &w, &name1);
            addEdge(index[name], index[name1], w);
        }

        //正图
        bool sol = spfa(0, n);
        if(sol)
            printf("Case %d: Yes\n", cas_c++);
        else
            printf("Case %d: No\n", cas_c++);
    }
    return 0;
}