codeforces-1194 (div2)

谁践踏了优雅 2023-10-11 13:26 19阅读 0赞

A.输出M \* 2

![ContractedBlock.gif][] ![ExpandedBlockStart.gif][]

B.记录每行每列的黑格子数量，然后N ^ 2枚举每个岔路口,选出最少的需要补全横竖两条线使得成功的方案

![ContractedBlock.gif][] ![ExpandedBlockStart.gif][]

C.首先判断s能不能变成t,即t中有没有一个子序列为s,然后再统计把s变为t需要的各个字母的数量，如果p中全部存在则YES

![ContractedBlock.gif][] ![ExpandedBlockStart.gif][]

#include <map>
    #include <set>
    #include <ctime>
    #include <cmath>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <string>
    #include <bitset>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <sstream>
    #include <iostream>
    #include <algorithm>
    #include <functional>
    using namespace std;
    #define For(i, x, y) for(int i=x;i<=y;i++)  
    #define _For(i, x, y) for(int i=x;i>=y;i--)
    #define Mem(f, x) memset(f,x,sizeof(f))  
    #define Sca(x) scanf("%d", &x)
    #define Sca2(x,y) scanf("%d%d",&x,&y)
    #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
    #define Scl(x) scanf("%lld",&x)  
    #define Pri(x) printf("%d\n", x)
    #define Prl(x) printf("%lld\n",x)  
    #define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
    #define LL long long
    #define ULL unsigned long long  
    #define mp make_pair
    #define PII pair<int,int>
    #define PIL pair<int,long long>
    #define PLL pair<long long,long long>
    #define pb push_back
    #define fi first
    #define se second 
    typedef vector<int> VI;
    int read(){
         int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){
         if (c == '-') f = -1;c = getchar();}
    while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
    const double PI = acos(-1.0);
    const double eps = 1e-9;
    const int maxn = 110;
    const int INF = 0x3f3f3f3f;
    const int mod = 1e9 + 7; 
    int N,M,K;
    char s[maxn],t[maxn],p[maxn];
    int vis[maxn];
    int main(){
        int T; Sca(T);
        while(T--){
            for(int i = 0 ; i < 26; i ++) vis[i] = 0;
            scanf("%s%s%s",s + 1,t + 1,p + 1);
            int l1 = strlen(s + 1),l2 = strlen(t + 1);
            if(l1 > l2){
                puts("NO"); continue;
            }    
            int cnt = 1;
            for(int i = 1; i <= l2; i ++){
                if(cnt <= l1 && t[i] == s[cnt]){
                    cnt++;
                }else{
                    vis[t[i] - 'a']++;
                }
            }
            if(cnt <= l1){
                puts("NO"); continue;
            }
            for(int i = 1; p[i]; i ++){
                vis[p[i] - 'a']--;
            }
            bool flag = 1;
            for(int i = 0; i < 26; i ++) if(vis[i] > 0) flag = 0;
            if(flag) puts("YES");
            else puts("NO");
        }
        return 0;
    }

D.先打一个SG表

![1468815-20190721214739644-1421072574.png][]

发现k不为3的倍数的时候为normal情况，每三个出现一个Bob

k为3的倍数的时候循环周期为(k / 3 - 1) \* 3 + 4,即(k / 3 - 1)个正常周期以及最后一个长度为4的非正常周期

![ContractedBlock.gif][] ![ExpandedBlockStart.gif][]

1.先将竖线按照x轴从小到大排列，将横线按照x2从小到大排列，然后n ^ 2枚举竖线

2.每次枚举，先将x1在左边竖线左边的横线全部记录,然后从左往右扫的过程中将x2在右边竖线左边的横线弹出，因为竖线的x坐标递增,所以弹出的横线对后续不会起到贡献。这样就处理完了x轴的大小关系

3.记录的过程用树状数组维护,将每次记录的横线y坐标计入树状数组，然后查询右边直线的时候查询y1 - y2的区间和就可以统计有多少横线v在这两条竖线中满足要求

最终这两条竖线给出的贡献是v \* (v - 1) / 2

时间复杂度n²logn

![ContractedBlock.gif][] ![ExpandedBlockStart.gif][]

#include <map>
    #include <set>
    #include <ctime>
    #include <cmath>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <string>
    #include <bitset>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <sstream>
    #include <iostream>
    #include <algorithm>
    #include <functional>
    using namespace std;
    #define For(i, x, y) for(int i=x;i<=y;i++)  
    #define _For(i, x, y) for(int i=x;i>=y;i--)
    #define Mem(f, x) memset(f,x,sizeof(f))  
    #define Sca(x) scanf("%d", &x)
    #define Sca2(x,y) scanf("%d%d",&x,&y)
    #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
    #define Scl(x) scanf("%lld",&x)  
    #define Pri(x) printf("%d\n", x)
    #define Prl(x) printf("%lld\n",x)  
    #define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
    #define LL long long
    #define ULL unsigned long long  
    #define mp make_pair
    #define PII pair<int,int>
    #define PIL pair<int,long long>
    #define PLL pair<long long,long long>
    #define pb push_back
    #define fi first
    #define se second 
    typedef vector<int> VI;
    int read(){
         int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){
         if (c == '-') f = -1;c = getchar();}
    while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
    const double PI = acos(-1.0);
    const double eps = 1e-9;
    const int maxn = 10010;
    const int INF = 0x3f3f3f3f;
    const int mod = 1e9 + 7; 
    int N,M,K;
    struct node{
        int x1,x2,y;
        node(int x1 = 0,int x2 = 0,int y = 0):x1(x1),x2(x2),y(y){}
    }row[maxn];
    struct node2{
        int y1,y2,x;
        node2(int y1 = 0,int y2 = 0,int x = 0):y1(y1),y2(y2),x(x){}
    }cul[maxn];
    bool cmp(node2 a,node2 b){
        return a.x < b.x;
    }
    bool cmp2(node a,node b){
        return a.x2 < b.x2;
    }
    int tree1[maxn],tree2[maxn];
    void add(int u,int v){
        for(;u < maxn; u += u & -u) tree1[u] += v;
    }
    LL get(int u){
        LL ans = 0;
        for(;u > 0; u -= u & -u) ans += tree1[u];
        return ans;
    }
    int main(){
        Sca(N); int cnt1 = 0,cnt2 = 0;
        for(int i = 1; i <= N ; i ++){
            int x1,y1,x2,y2;
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);y1 += 5001; y2 += 5001;
            if(x1 == x2){
                if(y1 > y2) swap(y1,y2);
                cul[++cnt1] = node2(y1,y2,x1);
            }else{
                if(x1 > x2) swap(x1,x2);
                row[++cnt2] = node(x1,x2,y1);
            }
        }
        LL ans = 0;
        sort(cul + 1,cul + 1 + cnt1,cmp);
        sort(row + 1,row + 1 + cnt2,cmp2);
        for(int i = 1; i <= cnt1; i ++){
            Mem(tree1,0); Mem(tree2,0);
            for(int j = 1; j <= cnt2; j ++){
                if(row[j].x1 <= cul[i].x && row[j].y >= cul[i].y1 && row[j].y <= cul[i].y2){
                    add(row[j].y,1);
                }
            }    
            int cnt = 1;
            for(int j = i + 1; j <= cnt1;j ++){
                while(cnt <= cnt2 && row[cnt].x2 < cul[j].x){
                    if(row[cnt].x1 <= cul[i].x && row[cnt].y >= cul[i].y1 && row[cnt].y <= cul[i].y2){
                        add(row[cnt].y,-1);
                    }
                    cnt++;
                }
                LL t = get(cul[j].y2) - get(cul[j].y1 - 1);
                ans += t * (t - 1) / 2;
            }
        }
        Prl(ans);
        return 0;
    }

转载于:https://www.cnblogs.com/Hugh-Locke/p/11222962.html

[ContractedBlock.gif]: https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif
[ExpandedBlockStart.gif]: https://images.cnblogs.com/OutliningIndicators/ExpandedBlockStart.gif
[1468815-20190721214739644-1421072574.png]: https://img2018.cnblogs.com/blog/1468815/201907/1468815-20190721214739644-1421072574.png