codeforces-1194 (div2)

A.输出M * 2

#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;
int main(){
    int T; Sca(T);
    while(T--){
        Sca2(N,M);
        Pri(M * 2);
    }
    return 0;
}
A

 

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

#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 = 5e4 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
string MAP[maxn];
LL row[maxn],cul[maxn];
int main(){
    int T; Sca(T);
    while(T--){
        Sca2(N,M);
        for(int i = 0 ; i < N ; i ++) row[i] = 0;
        for(int j = 0 ; j < M; j ++) cul[j] = 0;
        for(int i = 0; i < N ; i ++){
            cin >> MAP[i];
        }
        for(int i = 0; i < N ; i ++){
            for(int j = 0 ; j < M; j ++){
                if(MAP[i][j] == '*'){
                    row[i]++; cul[j]++;
                }
            }
        }
        LL ans = INF;
        for(int i = 0 ; i < N ; i ++){
            for(int j = 0 ; j < M ; j ++){
                ans = min(ans,N - row[i] + M - cul[j] - (MAP[i][j] == '.'));
            }
        }
        Prl(ans);
    }
    return 0;
}
B

 

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

#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;
}
C

 

D.先打一个SG表

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

#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; 
LL N,K;
int main(){
    int T; Sca(T);
    while(T--){
        scanf("%lld%lld",&N,&K);
        if(!N){
            puts("Bob"); continue;
        }
        if(K % 3){
            if(N % 3) puts("Alice");
            else puts("Bob");
        }else{
            LL t = (K / 3 - 1) * 3 + 4;
            N %= t;
            if(!N || (N != t - 1 && !(N % 3))) puts("Bob");
            else puts("Alice");
        }
    }
    return 0;
}
D

 

E.

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

#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;
}
E

 

posted @ 2019-07-21 21:54  Hugh_Locke  阅读(323)  评论(0编辑  收藏  举报