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POJ - 2240 - Humidex = SPFA / Floyd

http://poj.org/problem?id=2240

给定几种货币,和他们之间的汇率,求是不是有一个方法使得货币兑换一圈之后变多。

总而言之就是是否存在回路,路径上的权值的积大于1。

这个描述看起来就很SPFA,假如把路径的权值的积取负对数,就是问是否有负环。

直接上SPFA,果然很慢,要844ms(可能是常数不好)。

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;

const int MAXN = 35;
const int MAXM = 1005;

int top;
int head[MAXN];
struct Edge {
    int v, nxt;
    double w;
} edge[MAXM];

void init() {
    top = 0;
    memset(head, -1, sizeof(head));
}

void add_edge(int u, int v, double w) {
    ++top;
    edge[top].v = v;
    edge[top].w = w;
    edge[top].nxt = head[u];
    head[u] = top;
}

bool vis[MAXN];
int cnt[MAXN];
double dis[MAXN];

queue<int>q;
bool spfa(int s, int n) {
    memset(vis, 0, sizeof(vis));
    memset(cnt, 0, sizeof(cnt));
    for(int i = 0; i < MAXN; ++i)
        dis[i] = 1e18;

    while(!q.empty())
        q.pop();
    q.push(s);
    vis[s] = 1;
    cnt[s] = 1;
    dis[s] = 0;

    while(!q.empty()) {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; i != -1; i = edge[i].nxt) {
            int v = edge[i].v;
            if(dis[v] > dis[u] + edge[i].w) {
                dis[v] = dis[u] + edge[i].w;
                if(!vis[v]) {
                    vis[v] = 1;
                    q.push(v);
                    if(++cnt[v] > n)
                        return 0;
                }
            }
        }
    }
    return 1;
}

map<string, int> M;

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int n, m, ti = 0;
    while(~scanf("%d", &n)) {
        if(n == 0)
            break;
        M.clear();
        for(int i = 1; i <= n; ++i) {
            string s;
            cin >> s;
            M[s] = i;
        }
        scanf("%d", &m);
        init();
        while(m--) {
            string s;
            cin >> s;
            int u = M[s];
            double w;
            cin >> w;
            w = -log(w);
            cin >> s;
            int v = M[s];
            add_edge(u, v, w);
        }

        bool ans = 1;
        for(int i = 1; i <= n; ++i) {
            ans &= spfa(i, n);
            if(ans == 0)
                break;
        }

        printf("Case %d: ", ++ti);
        puts(ans == 0 ? "Yes" : "No");
    }
}

我记得SPFA也是单源最短路来的,会不会有更好的办法呢?

Floyd也可以,不过没有快多少。

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;

const int MAXN = 35;
double dis[MAXN][MAXN];

map<string, int> M;

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int n, m, ti = 0;
    while(~scanf("%d", &n)) {
        if(n == 0)
            break;
        M.clear();
        for(int i = 1; i <= n; ++i) {
            string s;
            cin >> s;
            M[s] = i;
        }
        scanf("%d", &m);
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= n; ++j) {
                dis[i][j] = 0;
                if(i == j)
                    dis[i][j] = 1;
            }
        }
        while(m--) {
            string s;
            cin >> s;
            int u = M[s];
            double w;
            cin >> w;
            cin >> s;
            int v = M[s];
            dis[u][v] = w;
        }

        for(int k = 1; k <= n; ++k) {
            for(int i = 1; i <= n; ++i) {
                for(int j = 1; j <= n; ++j)
                    dis[i][j] = max(dis[i][j], dis[i][k] * dis[k][j]);
            }
        }

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

        printf("Case %d: ", ++ti);
        puts(ans == 0 ? "Yes" : "No");
    }
}
posted @ 2019-10-21 12:10  Inko  阅读(...)  评论(...编辑  收藏