codeforces 723 A,B,C,D

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A - The New Year: Meeting Friends

水题，中位数。。。

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;int main(){    int a,b,c;    while(scanf("%d%d%d", &a, &b, &c) != EOF){        int ans1 = max(max(a, b), c);        int ans2 = min(min(a, b), c);        printf("%d\n", ans1 - ans2);    }    return 0;}

B - Text Document Analysis

模拟，细节问题注意一下即可，尤其是到了字符串结束的时候还需要进判断。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 300;char s[maxn];int main(){    int ans1, ans2, cnt, flag, n;    while(scanf("%d", &n) != EOF){        ans1 = ans2 = cnt = flag = 0;        scanf("%s", s);        for(int i = 0; i < n; i++){            if(s[i] == '_'){                if(!flag)                    ans1 = max(cnt, ans1);                else{                    if(cnt){        //                printf("i = %d\n", i);                        ans2++;                    }                }                cnt = 0;            }            else if(s[i] == '('){                ans1 = max(cnt, ans1);                cnt = 0;                flag = 1;            }            else if(s[i] == ')'){                if(cnt){          //          printf("i = %d\n", i);                    ans2++;                }                flag = 0;                cnt = 0;            }            else cnt++;        }        ans1 = max(ans1, cnt);        printf("%d %d\n", ans1, ans2);    }    return 0;}

C - Polycarp at the Radio

题意较难理解。其实（1，m）出现的最大次数应该为（n/m），这样才能保证使bj的最小值最大，然后模拟处理一下字符串即可。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 2005;int a[maxn], b[maxn], vis[maxn];int main(){    int n,m;    while(scanf("%d%d", &n, &m) != EOF){        memset(vis, 0, sizeof(vis));        int cnt = n/m;        for(int i = 0; i < n; i++){            scanf("%d", &a[i]);            if(a[i] <= m && a[i] >= 1) vis[a[i]]++;        }        int ans = 0;        for(int i = 1; i <= m; i++){            if(vis[i] < cnt) ans += (cnt - vis[i]);        }        printf("%d %d\n", cnt, ans);        int i, flag = 1;        for(i = 0; i < n; i++){            while(vis[flag] >= cnt && flag <= m) flag++;            if(flag > m) break;            if(a[i] >= 1 && a[i] <= m){   //             printf("vis[%d] = %d, flag = %d\n", a[i], vis[a[i]], flag);                if(vis[a[i]] > cnt){                    vis[a[i]]--;                    vis[flag]++;                    b[i] = flag;                }                else b[i] = a[i];            }            else{                vis[flag]++;                b[i] = flag;            }        }        for(; i < n; i++) b[i] = a[i];        int fi = 0;        for(int i = 0; i < n; i++){            if(!fi) printf("%d", b[i]), fi = 1;            else printf(" %d", b[i]);        }        printf("\n");    }    return 0;}

D - Lakes in Berland

bfs求连通块

#include <queue>#include <cstdio>#include <vector>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 55;char maze[maxn][maxn];int n, m, k, vis[maxn][maxn];bool flag;int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};struct Node{    int x, y;    Node(){}    Node(int a,int b):x(a), y(b){}};struct loc{    int x, y, val;    loc(int a,int b,int c):x(a), y(b), val(c){}    bool operator < (const loc temp)const{        return val < temp.val;    }};vector <loc> vec;int judge(int x,int y){    if(x == 0 || x == n-1 || y == 0 || y == m-1)        return false;    return true;}int bfs1(int x,int y){    flag = true;    int ans = 1;    queue <Node> que;    que.push(Node(x, y));    flag = flag & judge(x, y);    vis[x][y] = 1;    while(!que.empty()){        Node cur = que.front(); que.pop();        for(int i = 0; i < 4; i++){            int xx = cur.x + dir[i][0];            int yy = cur.y + dir[i][1];            if(xx < 0 || xx >= n || yy < 0 || yy >= m || vis[xx][yy]) continue;            if(maze[xx][yy] == '.'){                flag = flag&judge(xx, yy);                ans++;                vis[xx][yy] = 1;                que.push(Node(xx, yy));            }        }    }    return ans;}void bfs2(int x,int y){    queue <Node> q;    memset(vis, 0, sizeof(vis));    q.push(Node(x, y));    maze[x][y] = '*';    vis[x][y] = 1;    while(!q.empty()){        Node cur = q.front(); q.pop();        for(int i = 0; i < 4; i++){            int xx = cur.x + dir[i][0];            int yy = cur.y + dir[i][1];            if(xx < 0 || xx >= n || yy < 0 || yy >= m || vis[xx][yy]) continue;            if(maze[xx][yy] == '.'){                maze[xx][yy] = '*';                vis[xx][yy] = 1;                q.push(Node(xx, yy));            }        }    }}void solve(){    memset(vis, 0, sizeof(vis));    vec.clear();    for(int i = 0; i < n; i++){        for(int j = 0; j < m; j++){            if(maze[i][j] == '.' && !vis[i][j]){                int ans = bfs1(i, j);                if(flag) vec.push_back(loc(i, j, ans));            }        }    }    sort(vec.begin(), vec.end()); //   for(int i = 0; i < vec.size(); i++) printf("   %d", vec[i]);    int ret = 0;    for(int i = 0; i < vec.size()-k; i++){        ret += vec[i].val;        bfs2(vec[i].x, vec[i].y);    }    printf("%d\n", ret);    for(int i = 0; i < n; i++) printf("%s\n", maze[i]);}int main(){    while(scanf("%d%d%d", &n, &m, &k) != EOF){        for(int i = 0; i < n; i++) scanf("%s", maze[i]);        solve();    }    return 0;}

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