hdu 2389 Rain on your Parade （二分匹配 Hc 算法）

題目鏈鎖 http://acm.hdu.edu.cn/showproblem.php?pid=2389

先生成最短增廣路，在此基礎進行增廣，複雜度sqrt(n)*e

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<vector>
using namespace std;


#define inf 0x3f3f3f3f
#define eps 1e-9
#define mod 1000000007
#define FOR(i,s,t) for(int i = s; i < t; ++i )
#define REP(i,s,t) for( int i = s; i <= t; ++i )
#define LL long long
#define ULL unsigned long long
#define pii pair<int,int>
#define MP make_pair
#define lson id << 1 , l , m
#define rson id << 1 | 1 , m + 1 , r 
#define maxn ( 3000+10 )
#define maxe ( 50000+10 )


struct node {
    int x, y;
    int dis ( node ot ) {
        return ( x - ot.x ) * ( x - ot.x ) + ( y - ot.y ) * ( y - ot.y );
    }
}g[maxn], u[maxn];

int sp[maxn];
int G[maxn][maxn];

int mx[maxn], my[maxn], dx[maxn], dy[maxn];
int dis, nx, ny;
int Q[maxn*2];
bool vis[maxn];

bool search() {
    memset( dx, -1, sizeof( dx ) );
    memset( dy, -1, sizeof( dy ) );
    dis = inf;
    int head = 0, tail = 0;
    for( int i = 0; i < nx; ++i ) {
        if( mx[i] == -1 ) Q[tail ++] = i, dx[i] = 0;
    }
    while( head < tail ) {
        int u = Q[head++];
        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[tail++] = my[v];
            }
        }
    }
    return dis != inf;
}

bool dfs ( int u ) {
    for( int i = 0; i < ny; ++i ) if( G[u][i] && !vis[i] && dy[i] == dx[u] + 1 ) {
        vis[i] =1;
        if( my[i] != -1 && dis == dy[i] ) continue;
        if( my[i] == -1 || dfs( my[i] ) ) {
            my[i] = u;
            mx[u] = i;
            return 1;
        }
    }
    return 0;
}

int match() {
    int res = 0;
    memset( mx, -1, sizeof( mx) );
    memset( my, -1, sizeof( my ) );
    while( search() ) {
        memset( vis, 0, sizeof( vis ) );
        for( int i= 0; i < nx; ++i )
            if( mx[i] == -1 && dfs( i ) ) ++res;
    }
    return res;
}
int main () {
    int T;
    scanf("%d", &T );
    int cas = 1;
    for( cas = 1; cas <= T; ++cas ) {
        int n, m, t;
        scanf("%d%d", &t, &m ) ;
        for( int i = 0; i < m; ++i ) {
            scanf("%d%d%d", &g[i].x, &g[i].y, &sp[i] );
        }
        scanf("%d", &n );
        for( int i = 0; i < n; ++i ) {
            scanf("%d%d", &u[i].x, &u[i].y );
        }
        memset( G, 0, sizeof( G ) );
        for( int i = 0; i < m; ++i ) {
            for( int j = 0; j < n; ++j ) {
                if( g[i].dis( u[j] ) <= ( sp[i] * t ) * ( sp[i] * t ) )
                    G[i][j] = 1;
            }
        }
        nx = m, ny = n;
        printf("Scenario #%d:\n", cas );
        printf("%d\n\n", match() );
    }
}