HDU 2389 Rain on your Parade（H-C算法，二分图匹配）

来源：互联网发布：军人网络泄密案例解析编辑：程序博客网时间：2024/04/29 12:52
解题思路：
比较适合大数据的二分图匹配，用匈牙利算法会超时
#include <cstring>#include <algorithm>#include <cstdio>#include <queue>#include <iostream>using namespace std;const int MAXN = 3000 + 10;const int INF=1<<28;/************************************************/int g[MAXN][MAXN],Mx[MAXN],My[MAXN],Nx,Ny;int dx[MAXN],dy[MAXN],dis;bool vst[MAXN];bool searchP(){    queue<int>Q;    dis=INF;    memset(dx,-1,sizeof(dx));    memset(dy,-1,sizeof(dy));    for(int i=0;i<Nx;i++)        if(Mx[i]==-1)        {            Q.push(i);            dx[i]=0;        }    while(!Q.empty())    {        int u=Q.front();        Q.pop();        if(dx[u]>dis)  break;        for(int v=0;v<Ny;v++)            if(g[u][v]&&dy[v]==-1)            {                dy[v]=dx[u]+1;                if(My[v]==-1)  dis=dy[v];                else                {                    dx[My[v]]=dy[v]+1;                    Q.push(My[v]);                }            }    }    return dis!=INF;}bool DFS(int u){    for(int v=0;v<Ny;v++)       if(!vst[v]&&g[u][v]&&dy[v]==dx[u]+1)       {           vst[v]=1;           if(My[v]!=-1&&dy[v]==dis) continue;           if(My[v]==-1||DFS(My[v]))           {               My[v]=u;               Mx[u]=v;               return 1;           }       }    return 0;}int MaxMatch(){    int res=0;    memset(Mx,-1,sizeof(Mx));    memset(My,-1,sizeof(My));    while(searchP())    {        memset(vst,0,sizeof(vst));        for(int i=0;i<Nx;i++)          if(Mx[i]==-1&&DFS(i))  res++;    }    return res;}/*******************************************************/struct Point{    int x, y;}A[MAXN], B[MAXN];int V[MAXN];int d(Point a, Point b){    int x = a.x - b.x;    int y = a.y - b.y;    return x * x + y * y;}int main(){    int T, kcase = 1;    scanf("%d", &T);    while(T--)    {        int n, m, t;        scanf("%d", &t);        scanf("%d", &n);        for(int i=0;i<n;i++)        {            scanf("%d%d", &A[i].x, &A[i].y);            scanf("%d", &V[i]);        }        scanf("%d", &m);        for(int i=0;i<m;i++)            scanf("%d%d", &B[i].x, &B[i].y);        Nx = n, Ny = m;        memset(g, 0, sizeof(g));        for(int i=0;i<n;i++)        {            for(int j=0;j<m;j++)            {                if(d(A[i], B[j]) <= (V[i] * V[i] * t * t))                {                    g[i][j] = 1;                }            }        }        printf("Scenario #%d:\n", kcase++);        printf("%d\n\n", MaxMatch());    }    return 0;}
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