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  • SPFA/Dijkstra POJ 3159 Candies

    题目传送门

    题意:n个人发糖果,B 比 A 多 C的糖果,问最后第n个人比第一个人多多少的糖果

    分析:最短路,Dijkstra 优先队列优化可过,SPFA竟然要用栈,队列超时!

    代码:

    /************************************************
    * Author        :Running_Time
    * Created Time  :2015-9-1 19:18:52
    * File Name     :POJ_3159.cpp
     ************************************************/
    
    #include <cstdio>
    #include <algorithm>
    #include <iostream>
    #include <sstream>
    #include <cstring>
    #include <cmath>
    #include <string>
    #include <vector>
    #include <queue>
    #include <deque>
    #include <stack>
    #include <list>
    #include <map>
    #include <set>
    #include <bitset>
    #include <cstdlib>
    #include <ctime>
    using namespace std;
    
    #define lson l, mid, rt << 1
    #define rson mid + 1, r, rt << 1 | 1
    typedef long long ll;
    const int N = 3e4 + 10;
    const int E = 150000 + 10;
    const int INF = 0x3f3f3f3f;
    const int MOD = 1e9 + 7;
    struct Edge {
        int v, w, nex;
        Edge (int _v = 0, int _w = 0) : v (_v), w (_w) {}
        bool operator < (const Edge &r) const {
            return w > r.w;
        }
    }edge[E];
    int d[N];
    int head[N];
    int vis[N];
    int cnt[N];
    int n, m, e;
    
    void init(void) {
        memset (head, -1, sizeof (head));
        e = 0;
    }
    
    void Dijkstra(int s)    {
        memset (vis, false, sizeof (vis));
        for (int i=1; i<=n; ++i)    d[i] = INF;
        d[s] = 0;
        priority_queue<Edge> Q;  Q.push (Edge (s, 0));
        while (!Q.empty ()) {
            int u = Q.top ().v; Q.pop ();
            if (vis[u]) continue;
            vis[u] = true;
            for (int i=head[u]; ~i; i=edge[i].nex)  {
                int v = edge[i].v, w = edge[i].w;
                if (!vis[v] && d[v] > d[u] + w) {
                    d[v] = d[u] + w;    Q.push (Edge (v, d[v]));
                }
            }
        }
    }
    
    void SPFA(int s)    {
        memset (vis, false, sizeof (vis));
        memset (d, INF, sizeof (d));
        d[s] = 0; vis[s] = true;
        stack<int> S;   S.push (s);
        while (!S.empty ()) {
            int u = S.top (); S.pop ();
            vis[u] = false;
            for (int i=head[u]; ~i; i=edge[i].nex)  {
                int v = edge[i].v, w = edge[i].w;
                if (d[v] > d[u] + w)    {
                    d[v] = d[u] + w;
                    if (!vis[v])    {
                        vis[v] = true;  S.push (v);
                    }
                }
            }
        }
    }
    
    void add_edge(int u, int v, int w)  {
        edge[e].v = v, edge[e].w = w;
        edge[e].nex = head[u];  head[u] = e++;
    }
    
    int main(void)    {
        while (scanf ("%d%d", &n, &m) == 2) {
            init ();
            for (int u, v, w, i=1; i<=m; ++i)   {
                scanf ("%d%d%d", &u, &v, &w);
                add_edge (u, v, w);
            }
            //Dijkstra (1);
            SPFA (1);
            printf ("%d
    ", d[n]);
        }
    
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Running-Time/p/4776758.html
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