zoukankan html css js c++ java

POJ 2240 Arbitrage （求负环）

Arbitrage

题目链接：

http://acm.hust.edu.cn/vjudge/contest/122685#problem/I

Description

Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent. Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.

Input

The input will contain one or more test cases. Om the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible. Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n.

Output

For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No".

Sample Input

3 USDollar BritishPound FrenchFranc 3 USDollar 0.5 BritishPound BritishPound 10.0 FrenchFranc FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc
6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

Sample Output

Case 1: Yes Case 2: No

Hint

##题意：求货币经过一系列兑换操作后能否升值.
##题解：转化为图模型后就是求满足条件的环是否存在. 这里把求最短路时的加法改成乘法即可，结果就是是否存在环使得路径大于1. 以下分别用三种方法求： bellman-ford和floyd用时都较多，800+ms. spfa只需要90+ms, 不过需要用c++交，否则TLE.(真是神奇)
##代码： ####spaf法：94ms (必须用c++交，否则TLE) ``` cpp #include #include #include #include #include #include #include #include #include #include #define LL long long #define eps 1e-8 #define maxn 1100 #define inf 0x3f3f3f3f #define IN freopen("in.txt","r",stdin); using namespace std;

int m,n,k;
int edges, u[maxn], v[maxn];
double w[maxn];
int first[maxn], _next[maxn];
double dis[maxn];

void add_edge(int s, int t, double val) {
u[edges] = s; v[edges] = t; w[edges] = val;
_next[edges] = first[s];
first[s] = edges++;
}

queue q;
bool inq[maxn];
int inq_cnt[maxn];
bool spfa(int s) {
memset(inq, 0, sizeof(inq));
memset(inq_cnt, 0, sizeof(inq_cnt));
for(int i=1; i<=n; i++) dis[i] = 0; dis[s] = 1;
while(!q.empty()) q.pop();
q.push(s); inq_cnt[s]++;

while(!q.empty()) {
    int p = q.front(); q.pop();
    inq[p] = 0;
    for(int e=first[p]; e!=-1; e=_next[e]) {
        double tmp = dis[u[e]] * w[e];
        if(dis[v[e]] < tmp) {
            dis[v[e]] = tmp;
            if(!inq[v[e]]) {
                q.push(v[e]);
                inq[v[e]] = 1;
                inq_cnt[v[e]]++;
                if(inq_cnt[v[e]] >= n) return 0;
            }
        }
    }
}

return 1;

}

map<string,int> name;

int main(int argc, char const *argv[])
{
//IN;

int ca = 1;
while(scanf("%d", &n) != EOF && n)
{
    memset(first, -1, sizeof(first));
    edges = 0;
    name.clear();

    for(int i=1; i<=n; i++) {
        string s; cin >> s;
        name.insert(make_pair(s, i));
    }
    cin >> m;
    for(int i=1; i<=m; i++) {
        string s,t; double w;
        cin>> s >> w >> t;
        int u = name.find(s)->second;
        int v = name.find(t)->second;
        add_edge(u,v,w);
    }

    if(!spfa(1)) printf("Case %d: Yes
", ca++);
    else printf("Case %d: No
", ca++);
}

return 0;

}


####bellman-ford法：875ms
``` cpp
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define eps 1e-8
#define maxn 1100
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int m,n,k;
int edges, u[maxn], v[maxn];
double w[maxn];
int first[maxn], next[maxn];
double dis[maxn];

void add_edge(int s, int t, double val) {
    u[edges] = s; v[edges] = t; w[edges] = val;
    next[edges] = first[s];
    first[s] = edges++;
}

bool bellman(int s) {
    for(int i=1; i<=n; i++) dis[i]=0; dis[s] = 1;

    for(int i=1; i<=n; i++) {
        for(int e=0; e<edges; e++) {
            double tmp = dis[u[e]] * w[e];
            if(dis[v[e]] < tmp) {
                dis[v[e]] = dis[u[e]] * w[e];
                if(i == n) return 0;
            }
        }
    }

    return 1;
}

map<string,int> name;

int main(int argc, char const *argv[])
{
    //IN;

    int ca = 1;
    while(scanf("%d", &n) != EOF && n)
    {
        memset(first, -1, sizeof(first));
        edges = 0;
        name.clear();

        for(int i=1; i<=n; i++) {
            string s; cin >> s;
            name.insert(make_pair(s, i));
        }
        cin >> m;
        for(int i=1; i<=m; i++) {
            string s,t; double w;
            cin>> s >> w >> t;
            int u = name.find(s)->second;
            int v = name.find(t)->second;
            add_edge(u,v,w);
        }

        if(!bellman(1)) printf("Case %d: Yes
", ca++);
        else printf("Case %d: No
", ca++);
    }

    return 0;
}

floyd法：875ms

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define eps 1e-8
#define maxn 35
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int m,n,k;
double dis[maxn][maxn];

void floyd() {
    for(int k=1; k<=n; k++)
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
        if(dis[i][j] < dis[i][k]*dis[k][j])
            dis[i][j] = dis[i][k] * dis[k][j];
}

map<string,int> name;

int main(int argc, char const *argv[])
{
    //IN;

    int ca = 1;
    while(scanf("%d", &n) != EOF && n)
    {
        name.clear();
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                dis[i][j] = (i==j? 1.0:inf);

        for(int i=1; i<=n; i++) {
            string s; cin >> s;
            name.insert(make_pair(s, i));
        }
        cin >> m;
        for(int i=1; i<=m; i++) {
            string s,t; double w;
            cin>> s >> w >> t;
            int u = name.find(s)->second;
            int v = name.find(t)->second;
            dis[u][v] = w;
        }

        floyd();

        int flag = 1;
        for(int i=1; i<=n; i++)
            if(dis[i][i] > 1.0) {flag = 0;break;}

        if(!flag) printf("Case %d: Yes
", ca++);
        else printf("Case %d: No
", ca++);
    }

    return 0;
}

查看全文

相关阅读:
不常用函数总结
 高效update方案
 一次http完整的请求tcp报文分析
 类的初始化以及创建对象后的初始化
 [置顶] 编译背后的秘密
 html object元素
 JQuery初识
 Java多线程yield
智能电视TV开发---直播视频客户端结构设计和实现
 以Android环境为例的多线程学习笔记（二）-----------------锁和条件机制

原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5751517.html