• Rain on your Parade HDU


    在客人能够拿到的伞与客人之间建边  跑hc就好了。。。。

    看看别人的:https://blog.csdn.net/wall_f/article/details/8248350

    #include <iostream>
    #include <cstdlib>
    #include <cstdio>
    #include <cstring>
    #include <queue>
    #include <cmath>
    using namespace std;
     
    const int MAXN = 3010;
    const int MAXM = 3010*3010;
    const int INF = 0x3f3f3f3f;
     
    struct Edge
    {
        int v;
        int next;
    }edge[MAXM];
     
    struct node
    {
        double x, y;
        double v;
    }a[MAXN], b[MAXN];
     
    int nx, ny;
    int cnt;
    int t;
    int dis;
     
    int first[MAXN];
    int xlink[MAXN], ylink[MAXN];
    int dx[MAXN], dy[MAXN];
    int vis[MAXN];
     
    void init()
    {
        cnt = 0;
        memset(first, -1, sizeof(first));
        memset(xlink, -1, sizeof(xlink));
        memset(ylink, -1, sizeof(ylink));
    }
     
    void read_graph(int u, int v)
    {
        edge[cnt].v = v;
        edge[cnt].next = first[u], first[u] = cnt++;
    }
     
    int bfs()
    {
        queue<int> q;
        dis = INF;
        memset(dx, -1, sizeof(dx));
        memset(dy, -1, sizeof(dy));
        for(int i = 0; i < nx; i++)
        {
            if(xlink[i] == -1)
            {
                q.push(i);
                dx[i] = 0;
            }
        }
        while(!q.empty())
        {
            int u = q.front(); q.pop();
            if(dx[u] > dis) break;
            for(int e = first[u]; e != -1; e = edge[e].next)
            {
                int v = edge[e].v;
                if(dy[v] == -1)
                {
                    dy[v] = dx[u] + 1;
                    if(ylink[v] == -1) dis = dy[v];
                    else
                    {
                        dx[ylink[v]] = dy[v]+1;
                        q.push(ylink[v]);
                    }
                }
            }
        }
        return dis != INF;
    }
     
    int find(int u)
    {
        for(int e = first[u]; e != -1; e = edge[e].next)
        {
            int v = edge[e].v;
            if(!vis[v] && dy[v] == dx[u]+1)
            {
                vis[v] = 1;
                if(ylink[v] != -1 && dy[v] == dis) continue;
                if(ylink[v] == -1 || find(ylink[v]))
                {
                    xlink[u] = v, ylink[v] = u;
                    return 1;
                }
            }
        }
        return 0;
    }
     
    int MaxMatch()
    {
        int ans = 0;
        while(bfs())
        {
            memset(vis, 0, sizeof(vis));
            for(int i = 0; i < nx; i++) if(xlink[i] == -1)
            {
                ans += find(i);
            }
        }
        return ans;
    }
     
    /*double dist(const node a, const node b) //TLE,无力吐槽了
    {
        return sqrt(pow((a.x-b.x), 2.0) + pow((a.y-b.y), 2.0));
    }*/
     
    double dist(const node a, const node b)
    {
        return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
    }
     
    void read_case()
    {
        init();
        int Time;
        scanf("%d", &Time);
        scanf("%d", &nx);
        for(int i = 0; i < nx; i++)
        {
            scanf("%lf%lf%lf", &a[i].x, &a[i].y, &a[i].v);
        }
        scanf("%d", &ny);
        for(int i = 0; i < ny; i++)
        {
            scanf("%lf%lf", &b[i].x, &b[i].y);
        }
        for(int i = 0; i < nx; i++)
        {
            for(int j = 0; j < ny; j++)
            {
                double limit = a[i].v*Time;
                double s = dist(a[i], b[j]);
                if(s <= limit) read_graph(i, j);
            }
        }
    }
     
    void solve()
    {
        read_case();
        int ans = MaxMatch();
        printf("%d
    
    ", ans); //注意格式 
    }
     
    int main()
    {
        int T, times = 0;
        scanf("%d", &T);
        while(T--)
        {
            printf("Scenario #%d:
    ", ++times);
            solve();
        }
        return 0;
    }
    自己选择的路,跪着也要走完。朋友们,虽然这个世界日益浮躁起来,只要能够为了当时纯粹的梦想和感动坚持努力下去,不管其它人怎么样,我们也能够保持自己的本色走下去。
  • 相关阅读:
    张量自动求导和计算图
    34 深入浅出 x86 中断机制(下)
    33 深入浅出 x86 中断机制(上)
    使用STM32 PWM控制LCD背光
    32 获取物理内存容量 下
    ucGUI内存管理相关
    STM32 SPI双机通信参考
    IAR_UCOSII_UCOSIII
    每日总结
    每日总结
  • 原文地址:https://www.cnblogs.com/WTSRUVF/p/9310650.html
Copyright © 2020-2023  润新知