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  • 2017 计蒜之道 初赛 第四场

    A题
    贪心+dfs
    dfs求出每个分割后小块中小孔的个数,除2是能安装芯片的个数,累加即可

    #include <bits/stdc++.h>
    #define nmax 105
    using namespace std;
    int mp[nmax][nmax];
    int sx[] = {0,1,0,-1};
    int sy[] = {1,0,-1,0};
    int n,m,k;
    bool check(int x, int y)
    {
        if(x<0 || x>=n || y<0|| y>=m || mp[x][y] == 1){
            return false;
        }else{
            return true;
        }
    }
    int dfs(int x, int y,int depth)
    {
        mp[x][y] = 1;
        for(int i = 0; i<4;++i){
            int spx = x+sx[i];
            int spy = y+sy[i];
            if(check(spx,spy)){
                return dfs(spx,spy,depth+1);
            }
        }
        return depth;
    
    }
    void out()
    {
        for(int i = 0; i<n;++i){
            for(int j = 0; j<m;++j){
                printf("%d ",mp[i][j]);
            }
            printf("
    ");
        }
    }
    int main()
    {
        while(scanf("%d %d %d",&n,&m,&k) != EOF){
            memset(mp,0,sizeof(mp));
            int d, c;
            for(int i = 0; i<k;++i){
                scanf("%d %d",&d,&c);
                if(d == 0){
                    for(int j = 0; j<m;++j){
                        mp[c-1][j] = 1;
                    }
                }else{
                    for(int j = 0; j<n;++j){
                        mp[j][c-1] = 1;
                    }
                }
            }
            int temp = 0,ans = 0;
            for(int i = 0; i<n;++i){
                for(int j = 0; j<m;++j){
                    if(check(i,j)){
                        temp = dfs(i,j,1);
                        ans +=temp /2;
                    }
                }
            }
            printf("%d
    ",ans);
        }
        return 0;
    }

    B题
    map存pair和个数,然后o(n)扫描map即可

    #include <bits/stdc++.h>
    using namespace std;
    map<pair<int,int>, int> mp;
    pair<int,int> p;
    int main()
    {
        int n;
        while(scanf("%d",&n) != EOF){
            mp.clear();
            for(int i = 0; i<n;++i){
                int x1,x2,y1,y2;
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                int dx = x2-x1;
                int dy = y2-y1;
                p = make_pair(dx,dy);
                mp[p]++;
            }
            map<pair<int,int>,int>::iterator it;
            int nowbiggest = -1e9;
            for(it = mp.begin();it!=mp.end();it++){
                if(it->second>nowbiggest){
                    nowbiggest = it->second;
                    p = it->first;
                }
            }
            printf("%d %d
    ",p.first,p. second);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/pengwill/p/7367079.html
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