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  • Oil Skimming HDU

    Oil Skimming

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3426    Accepted Submission(s): 1432


    Problem Description
    Thanks to a certain "green" resources company, there is a new profitable industry of oil skimming. There are large slicks of crude oil floating in the Gulf of Mexico just waiting to be scooped up by enterprising oil barons. One such oil baron has a special plane that can skim the surface of the water collecting oil on the water's surface. However, each scoop covers a 10m by 20m rectangle (going either east/west or north/south). It also requires that the rectangle be completely covered in oil, otherwise the product is contaminated by pure ocean water and thus unprofitable! Given a map of an oil slick, the oil baron would like you to compute the maximum number of scoops that may be extracted. The map is an NxN grid where each cell represents a 10m square of water, and each cell is marked as either being covered in oil or pure water.
     
    Input
    The input starts with an integer K (1 <= K <= 100) indicating the number of cases. Each case starts with an integer N (1 <= N <= 600) indicating the size of the square grid. Each of the following N lines contains N characters that represent the cells of a row in the grid. A character of '#' represents an oily cell, and a character of '.' represents a pure water cell.
     
    Output
    For each case, one line should be produced, formatted exactly as follows: "Case X: M" where X is the case number (starting from 1) and M is the maximum number of scoops of oil that may be extracted.
     
    Sample Input
    1 6 ...... .##... .##... ....#. ....## ......
     
    Sample Output
    Case 1: 3
     
    Source
     
    Recommend
    lcy
     

    就是一个板题。。。

    相邻的两个#建边。。。。然后求最大匹配就好了

    用匈牙利就够了  我用的hc

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #define mem(a, b) memset(a, b, sizeof(a))
    using namespace std;
    const int maxn = 10010, INF = 0x7fffffff;
    int dx[maxn], dy[maxn], cx[maxn], cy[maxn], used[maxn];
    int nx, ny, dis, n;
    char str[610][610];
    int gra[610][610];
    vector<int> G[40005];
    int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
    int bfs()
    {
        queue<int> Q;
        dis = INF;
        mem(dx, -1);
        mem(dy, -1);
        for(int i=1; i<=nx; i++)
        {
            if(cx[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<G[u].size(); v++)
            {
                int i=G[u][v];
                if(dy[i] == -1)
                {
                    dy[i] = dx[u] + 1;
                    if(cy[i] == -1) dis = dy[i];
                    else
                    {
                        dx[cy[i]] = dy[i] + 1;
                        Q.push(cy[i]);
                    }
                }
            }
        }
        return dis != INF;
    }
    
    int dfs(int u)
    {
        for(int v=0; v<G[u].size(); v++)
        {
            int i = G[u][v];
            if(!used[i] && dy[i] == dx[u] + 1)
            {
                used[i] = 1;
                if(cy[i] != -1 && dis == dy[i]) continue;
                if(cy[i] == -1 || dfs(cy[i]))
                {
                    cy[i] = u;
                    cx[u] = i;
                    return 1;
                }
            }
        }
        return 0;
    }
    
    int hk()
    {
        int res = 0;
        mem(cx, -1);
        mem(cy, -1);
        while(bfs())
        {
            mem(used, 0);
            for(int i=1; i<=nx; i++)
                if(cx[i] == -1 && dfs(i))
                    res++;
        }
        return res;
    }
    
    
    int main()
    {
        int T, kase = 0;
        cin>> T;
        while(T--)
        {
            mem(gra, 0);
            int ans = 0;
            for(int i=0; i<maxn; i++) G[i].clear();
            cin>> n;
            for(int i=0; i<n; i++)
            {
                cin>> str[i];
                for(int j=0; j<n; j++)
                {
                    if(str[i][j] == '#')
                        gra[i][j] = ++ans;
    
                }
    
            }
            for(int i=0; i<n; i++)
            {
                for(int j=0; j<n; j++)
                {
                    if(str[i][j] == '#')
                        for(int k=0; k<4; k++)
                        {
                            int nx = i + dir[k][0];
                            int ny = j + dir[k][1];
                            if(str[nx][ny] == '#' && nx >= 0 && ny >= 0 && nx < n && ny < n)
                                G[gra[i][j]].push_back(gra[nx][ny]), G[gra[nx][ny]].push_back(gra[i][j]);
                        }
                }
            }
            nx = ny = ans;
            printf("Case %d: %d
    ",++kase, hk()/2);
        }
    
        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。
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  • 原文地址:https://www.cnblogs.com/WTSRUVF/p/9310769.html
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