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 105using 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("\n");    }}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\n",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\n",p.first,p. second);    }    return 0;}