Arbitrage(最短路问题)

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

0

Case 1: Yes
Case 2: No

题解: 判断是否有大于1的环,用floryd算法,判断大于1的环;

AC代码:
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-9

using namespace std;
typedef long long ll;
typedef pair<int,int>P;
const int M = 200;
const int INF = 0x3f3f3f3f;
const int mod = 10000007;
double dp[M][M];
string s;
map<string,int>mp;

int main(){
    int n,cnt = 0;
    while(cin>>n && n){
        mp.clear();
        for(int i = 1; i <= n; i++){
            cin>>s;
            mp[s] = i;
        }
        int m;
        cin>>m;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                if(i == j) dp[i][j] = 1;
                else dp[i][j] = 0;
            }
        }
        while(m--){
            int u,v;
            double c;
            cin>>s;
            u = mp[s];
            scanf("%lf",&c);
            cin>>s;
            v = mp[s];
            dp[u][v] = max(dp[u][v],c);
        }
        for(int k = 1; k <= n; k++){
            for(int i = 1; i <= n; i++){
                for(int j = 1; j <= n; j++){
                    dp[i][j] = max(dp[i][j],dp[i][k] * dp[k][j]);
                }
            }
        }
        int vis = 0;
        for(int i = 1; i <= n && !vis; i++){
            if(dp[i][i] > 1) vis = 1;
        }
        printf("Case %d: ",++cnt);
        if(vis) puts("Yes");
        else puts("No");
    }
    return 0;
}