codeforces round #648 (div2)总结

2020.6.8
最近连续两场翻车让我不得不加紧了一下训练。前几天没再划水了,昨天上午去医院看完奶奶下午回来锻炼完就坐在电脑前刷题,把二分做了做,重点练了下手动模拟,还是蛮重要的,因为前几场就是吃了手动模拟的亏,手速太慢导致明明ac/wa/tle情况一样别人涨分我掉分的情况。到了八点买了杯coco就坐在电脑前等待开始。昨天心态有点小小的波动,因为还是太曲折了。看这个结果就不难想到为什么了hhhh
codeforces round #648 (div2)总结_第1张图片

A题果不其然又是开幕雷击,题目很好想,就是找到行和列同时全都为0的点的数量,若是奇数则先手必胜,偶数则必败,然后wa2,淦。窝自己看了好一会才发现我把col[]加了两遍,然而row根本没动,就这还把样例水过去了也是太惨。结果就是改了之后秒了A。一起打的大三的群友估计对我无语了吧,什么菜鸡a题都能卡。

B题很有意思,问的是数组a内的元素分为0和1两种,只有种类不同才能合法互换,问能不能通过无限次数合法的互换将数列完全排好序。思路是不可能的,这辈子都不可能的。一开始以为是独立集排序然后比较端点大小。后来窝发现好像没考虑全情况,样例都过不了,朋友wa4,这可咋办呢。然后我看着数据手动模拟了下,有一种感觉是除了一开始升序的情况直接可以判断之外,其余只要同时有0和1同时存在,那么就一定可以排好序,但窝不会证明。样例过了就交了一发,然后re了点4,后来发现是数组没开够,开大之后一发过了???刚才被我微信自言自语烦死的老哥都懵了。事后我发邮件问了下Piot教授,Piot说他觉得对,但是没法证明。

然后看C,C是问a和b序列都是只包含1-n中的数字,每个数字出现一次,可以把a的元素集体左移或者右移,问a通过任意移动之后和b数组同位置元素相同的最大数量是多少。窝一开始没看见1-n每个数字只出现一次,写了个假算法跟朋友说我算法fAke了,不能处理重复数字。然后朋友跟我说,喵的每个数字本来就出现一次好吧?窝:???看到n可以开数组,然后这样就简单了,和之前一道dp类似。那么直接预处理a的元素对应的位置,然后算和b中元素的下标差值,开个桶投票,最多的一个即为答案。窝因为没查0和比n大的元素各wa了一次。后来想想边跑边统计难道不行么??就A了,窝太菜了。然后隔壁老哥写到C估计觉得这人好没意思,拿群友当聊天垃圾桶和工具人了么?

D有意思,问能不能在矩阵中空地(".")添加墙壁,让"B"代表的坏人不能到达(n,m),好人(“G”)可以。然后窝一开始就觉得首先要逃出去坏人不能和好人相邻,然后wa3。朋友告诉我我忘了在坏人周围筑墙,就是把坏人旁边的空地换成墙,然后在终点bfs看能不能收集齐所有的好人并且坏人不能和好人相邻。之后因为judge函数写串了差点没交上hhhh

E我觉得挺简单,但当时前面耽误时间太多了,没能开,呜呜呜呜。陪我一起打的老哥估计被我大话痨给折磨惨了。基本都是我自己自言自语一堆,然后莫名其妙就AC了。道个歉。不过也证明了训练的重要性,以前划水太厉害了,导致之前思维有所下降。还是得加强训练。po代码

A题:

#include 
using namespace std;
#define limit (100 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair
#define rep(i, a, b) for(int i = a; i <= b ; ++i)
#define per(i, a, b) for(int i = b ; i >= a ; --i)
#define mint(a,b,c) min(min(a,b), c)
#define MOD 998244353
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
typedef long long ll;
typedef unsigned long long ull;
ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = x * 10 + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int n ,k,kase;
int f[limit][limit],row[limit],col[limit];
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    cin>>kase;
    while(kase--){
        cin>>n>>k;
        int tot = 0;
        rep(i ,1, n){
            rep(j ,1, k){
                int val;
                cin>>val;
                if(val)f[i][j] = 1;
                else f[i][j] = 0;
            }
        }
        rep(j ,1, k){
            int flag = 0;
            rep(i ,1, n){
                if(f[i][j])flag = 1;
            }
            col[j] = flag;
        }
        rep(i ,1, n){
            int flag = 0;
            rep(j ,1, k){
                if(f[i][j])flag = 1;
            }
            row[i] = flag;
        }
        int ans2 = 0;
        rep(i ,1, n){
            if(!row[i])++tot;
        }
        rep(i ,1, k){
            if(!col[i])++ans2;
        }
        tot = min(tot, ans2);
        puts(tot % 2 == 1 ? "Ashish" : "Vivek");
    }

    return 0;
}

B题:

#include 
using namespace std;
#define limit (100000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair
#define rep(i, a, b) for(int i = a; i <= b ; ++i)
#define per(i, a, b) for(int i = b ; i >= a ; --i)
#define mint(a,b,c) min(min(a,b), c)
#define MOD 998244353
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
typedef long long ll;
typedef unsigned long long ull;
ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = x * 10 + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int n ,k,kase;
struct node{
    int at, bt;
}a[limit];
int g[limit] , f[limit];
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    cin>>kase;
    while(kase--){
        cin>>n;
        int cnt = 0 , cnt2 = 0;
        int last = -INF;
        int flag = 0;
        rep(i ,1, n){
            cin>>a[i].bt;
            if(a[i].bt < last)flag = 1;
            last = a[i].bt;
        }
        rep(i , 1, n){
            cin>>a[i].bt;
        }
        if(!flag){
            cout<<"Yes"<<endl;
            continue;
        }
        rep(i ,1, n){
            int s = a[i].at , t = a[i].bt;
            if(t){
                f[++cnt] = s;
            }else{
                g[++cnt2] = s;
            }
        }
        puts(min(cnt2, cnt) ? "Yes" : "No");
    }

    return 0;
}

C题:

#include 
using namespace std;
#define limit (1000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair
#define rep(i, a, b) for(int i = a; i <= b ; ++i)
#define per(i, a, b) for(int i = b ; i >= a ; --i)
#define mint(a,b,c) min(min(a,b), c)
#define MOD 998244353
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
typedef long long ll;
typedef unsigned long long ull;
ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = x * 10 + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int n ,k,kase;
int a[limit] , b[limit];
int posa[limit], dist[limit];
int bucket[limit];
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    cin>>n;
    rep(i ,1, n){
        cin>>a[i];
        posa[a[i]] = i;//反映射
    }
    rep(i ,1, n){
        cin>>b[i];
        if(posa[b[i]] < i){
            bucket[(posa[b[i]] + n - i)]++;
        }else{
            bucket[(posa[b[i]] - i)]++;
        }
    }
    int maxx = 0;
    rep(i ,0, 2 * n){
        maxx = max(maxx, bucket[i]);
    }
    cout<<maxx<<endl;
    return 0;
}

D题:

#include 
using namespace std;
#define limit (100 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair
#define rep(i, a, b) for(int i = a; i <= b ; ++i)
#define per(i, a, b) for(int i = b ; i >= a ; --i)
#define mint(a,b,c) min(min(a,b), c)
#define MOD 998244353
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
typedef long long ll;
typedef unsigned long long ull;
ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = x * 10 + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int n ,k,kase;
char mapp[limit][limit];
int vis[limit][limit];
bool legal(int x, int y){
    return x >= 1 && y >= 1 && x <=n && y <= k && mapp[x][y] != 'G' && mapp[x][y] != 'B'&& !vis[x][y] ;
}
bool legal2(int x, int y){
    return x >= 1 && y >= 1 && x <=n && y <= k && !vis[x][y] && mapp[x][y] != '#';
}

int main() {
#ifdef LOCAL
    FOPEN;
#endif
    cin>>kase;
    while(kase--){
        cin>>n>>k;
        int flag = 0;
        memset(vis, 0 , sizeof(vis));
        rep(i ,1, n){
            string str;
            cin>>str;
            rep(j ,1, k){
                mapp[i][j] = str[j - 1];
                if(mapp[i][j] == 'G')flag++;
            }
        }

        if(!flag){
            puts("Yes");
            continue;
        }
        int status = 0;
        rep(i ,1, n){
            rep(j , 1, k){
                if(mapp[i][j] == 'B'){
                    if(legal(i + 1, j)){
                        mapp[i+1][j] = '#';
                    }
                    if(legal(i - 1, j)){
                        mapp[i-1][j] = '#';
                    }
                    if(legal(i, j + 1)){
                        mapp[i][j+1] = '#';
                    }
                    if(legal(i, j - 1)){
                        mapp[i][j-1] = '#';
                    }
                }
            }
        }
        queue<pi(int, int)>q;
        vis[n][k] = 1;
        q.push({n,k});
        int t = 0;
        while(q.size()){
            auto it = q.front();
            q.pop();
            int i = it.first, j = it.second;
            if(legal2(i + 1, j)){
                vis[i+1][j] = 1;
                if(mapp[i+1][j] == 'G'){
                    ++t;
                }
                if(mapp[i+1][j] == 'B'){
                    status = 1;
                }
                q.push({i+1,j});
            }
            if(legal2(i - 1, j)){
                vis[i-1][j] = 1;
                if(mapp[i-1][j] == 'G'){
                    ++t;
                }
                if(mapp[i-1][j] == 'B'){
                    status = 1;
                }
                q.push({i-1,j});
            }
            if(legal2(i, j + 1)){
                vis[i][j+1] = 1;
                if(mapp[i][j+1] == 'G'){
                    ++t;
                }
                if(mapp[i][j+1] == 'B'){
                    status = 1;
                }
                q.push({i,j+1});
            }
            if(legal2(i, j - 1)){
                vis[i][j-1] = 1;
                if(mapp[i][j-1] == 'G'){
                    ++t;
                }
                if(mapp[i][j-1] == 'B'){
                    status = 1;
                }
                q.push({i,j-1});
            }
        }
        if(status){
            puts("No");
            continue;
        }
        if(t == flag){
            puts("Yes");
        }else{
            puts("No");
        }
    }

    return 0;
}

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