现在的位置: 首页 > 综合 > 正文

HDU 2426 Interesting Housing Problem

2019年04月06日 ⁄ 综合 ⁄ 共 1727字 ⁄ 字号 评论关闭

7月冒泡第5题。

由于为负数的房间不能入住,所以不需要读入,而直接初始化为-INF.这是一个很重要的初始化。因为,如果不这样做的话,可能学生可以一一对应一些房间。那么当计算是否完备匹配时,是比较麻烦的。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
using namespace std;

const int maxn = 510;
const int INF = 0x3f3f3f3f;

int n, m;

int W[maxn][maxn];
int Lx[maxn], Ly[maxn];

int Left[maxn];
bool S[maxn], T[maxn];

bool match(int i)
{
    S[i] = 1;
    for(int j = 1; j <= m; j++) if(Lx[i]+Ly[j] == W[i][j] && !T[j])
    {
        T[j] = 1;
        if(!Left[j] || match(Left[j]))
        {
            Left[j] = i;
            return 1;
        }
    }
    return 0;
}

void update()
{
    int a = INF;
    for(int i = 1; i <= n; i++) if(S[i])
    for(int j = 1; j <= m; j++) if(!T[j])
        a = min(a, Lx[i]+Ly[j]-W[i][j]);
        
    for(int i = 1; i <= n; i++)
    {
        if(S[i]) Lx[i] -= a;
    }
    for(int j = 1; j <= m; j++)
    {
        if(T[j]) Ly[j] += a;
    }
}

void KM()
{
    memset(Left, 0, sizeof(Left));
    memset(Lx, 0, sizeof(Lx));
    memset(Ly, 0, sizeof(Ly));
    
    for(int i = 1; i <= n; i++)
    {
        Left[i] = Lx[i] = Ly[i] = 0;
        for(int j = 1; j <= m; j++)
            Lx[i] = max(Lx[i], W[i][j]);
    }
    for(int i = 1; i <= n; i++)
    {
        for(;;)
        {
            for(int j = 1; j <= m; j++) S[j] = T[j] = 0;
            if(match(i)) break; else update();
        }
    }
}

int check()
{
    int ans = 0, count = 0;
    for(int i = 1; i <= m; i++) if(Left[i] != 0 && W[Left[i]][i] != -INF)
    {
        ans += W[Left[i]][i];
        count++;
    }
    if(count != n) return -1;
    return ans;
}

inline void readint(int &x)
{
    char c;
    c = getchar();
    while(!isdigit(c)) c = getchar();
    
    x = 0;
    while(isdigit(c)) x = x*10+c-'0', c = getchar();
}

inline void writeint(int x)
{
    if(x > 9) writeint(x/10);
    putchar(x%10+'0');
}

int E;

void read_case()
{
    
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= m; j++) W[i][j] = -INF;
    
    while(E--)
    {
        int x, y, z;
        scanf("%d%d%d", &x, &y, &z); x++; y++;
        if(z >= 0) W[x][y] = z;
    }
}

int times;

void solve()
{
    read_case();
    if(E == 0) { printf("-1\n"); return ;}
    if(n > m) { printf("-1\n"); return ; }
    KM();
    
    int ans = check();
    if(ans == -1) { printf("-1\n"); return ;}
    
    printf("Case %d: %d\n", ++times, ans);
}

int main()
{
    times = 0;
    while(~scanf("%d%d%d", &n, &m, &E))
    {
        solve();
    }
    return 0;
}

抱歉!评论已关闭.