牛客小白月赛6
https://www.nowcoder.com/acm/contest/136#question
所有代码：
https://pan.baidu.com/s/1B0a4CZ6HFev0iwDdtDwBcA
A.
好题
C.
树最长链
两种方法
1 /** 2 同样的遍历，若无特殊要求，用bfs！ 3 dfs要保存、返回函数中间状态， 4 dfs比bfs慢不少，如一棵有很多分叉的树。 5 6 题解的方法 7 */ 8 9 #include <cstdio> 10 #include <cstdlib> 11 #include <cmath> 12 #include <cstring> 13 #include <time.h> 14 #include <string> 15 #include <set> 16 #include <map> 17 #include <list> 18 #include <stack> 19 #include <queue> 20 #include <vector> 21 #include <bitset> 22 #include <ext/rope> 23 #include <algorithm> 24 #include <iostream> 25 using namespace std; 26 #define ll long long 27 #define minv 1e-6 28 #define inf 1e9 29 #define pi 3.1415926536 30 #define E 2.7182818284 31 const ll mod=1e9+7;//998244353 32 const int maxn=1e6+10; 33 34 struct node 35 { 36 int d; 37 node* next; 38 }*e[maxn],*p; 39 40 int q[maxn],l[maxn]; 41 bool vis[maxn]; 42 int head,tail; 43 44 void bfs(int d) 45 { 46 head=0,tail=1; 47 q[1]=d,l[1]=1; 48 memset(vis,0,sizeof(vis)); 49 while (head<tail) 50 { 51 head++; 52 p=e[ q[head] ]; 53 while (p) 54 { 55 if (!vis[p->d]) 56 { 57 vis[p->d]=1; 58 tail++; 59 q[tail]=p->d; 60 l[tail]=l[head]+1; 61 } 62 p=p->next; 63 } 64 } 65 } 66 67 int main() 68 { 69 int n,i,x,y; 70 scanf("%d",&n); 71 for (i=1;i<=n;i++) 72 e[i]=NULL; 73 for (i=1;i<n;i++) 74 { 75 scanf("%d%d",&x,&y); 76 p=(node*) malloc (sizeof(node*)); 77 p->d=y; 78 p->next=e[x]; 79 e[x]=p; 80 81 p=(node*) malloc (sizeof(node*)); 82 p->d=x; 83 p->next=e[y]; 84 e[y]=p; 85 } 86 bfs(1); 87 bfs(q[tail]); 88 printf("%d",l[tail]); 89 90 return 0; 91 }
1 /* 2 dp 3 */ 4 #include <cstdio> 5 #include <cstdlib> 6 #include <cmath> 7 #include <cstring> 8 #include <time.h> 9 #include <string> 10 #include <set> 11 #include <map> 12 #include <list> 13 #include <stack> 14 #include <queue> 15 #include <vector> 16 #include <bitset> 17 #include <ext/rope> 18 #include <algorithm> 19 #include <iostream> 20 using namespace std; 21 #define ll long long 22 #define minv 1e-6 23 #define inf 1e9 24 #define pi 3.1415926536 25 #define E 2.7182818284 26 const ll mod=1e9+7;//998244353 27 const int maxn=1000000+10; 28 29 //用vector比用node稍微慢一点，但是也许vector更方便 30 31 struct node 32 { 33 int d; 34 node* next; 35 }*point[maxn]; 36 37 //vector<int>e[maxn]; 38 bool vis[maxn]={0}; 39 int r; 40 41 //返回的是目前以d为根节点(之后会变)的最长链 42 int dfs(int d) 43 { 44 int value=0,v1=0,v2=0,v; 45 int dd; 46 node* p; 47 // vector<int>::iterator i; 48 vis[d]=1; 49 // for (i=e[d].begin();i!=e[d].end();i++) 50 p=point[d]; 51 while (p) 52 { 53 // dd=*i; 54 dd=p->d; 55 if (!vis[dd]) 56 { 57 v=dfs(dd); 58 value=max(value,v); 59 if (v>v1) 60 { 61 v2=max(v1,v2); 62 v1=v; 63 } 64 else if (v>v2) 65 v2=v; 66 } 67 p=p->next; 68 } 69 //d为根节点，左右各连接一条链，即求以d儿子为根节点的最长的两条链 70 r=max(r,v1+v2+1); 71 return value+1; 72 } 73 74 int main() 75 { 76 int n,x,y,i; 77 node* p; 78 scanf("%d",&n); 79 for (i=1;i<=n;i++) 80 point[i]=NULL; 81 for (i=1;i<n;i++) 82 { 83 scanf("%d%d",&x,&y); 84 p=(struct node *) malloc (sizeof(node)); 85 p->d=y; 86 p->next=point[x]; 87 point[x]=p; 88 89 p=(struct node *) malloc (sizeof(node)); 90 p->d=x; 91 p->next=point[y]; 92 point[y]=p; 93 // e[x].push_back(y); 94 // e[y].push_back(x); 95 } 96 i=dfs(1); 97 printf("%d",r); 98 return 0; 99 } 100 /* 101 4 102 1 2 103 1 3 104 1 4 105 */
vector
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <time.h> 6 #include <string> 7 #include <set> 8 #include <map> 9 #include <list> 10 #include <stack> 11 #include <queue> 12 #include <vector> 13 #include <bitset> 14 #include <ext/rope> 15 #include <algorithm> 16 #include <iostream> 17 using namespace std; 18 #define ll long long 19 #define minv 1e-6 20 #define inf 1e9 21 #define pi 3.1415926536 22 #define E 2.7182818284 23 const ll mod=1e9+7;//998244353 24 const int maxn=1e6+10; 25 26 vector<int>st; 27 char mode[10]; 28 29 /* 30 注意对于存在的指纹，相邻指纹距离>k 31 对于一个数x，只有指纹中大于等于x的第一个数y 和 该数y的上一个数z 才有可以与x的距离小于等于k 32 */ 33 34 int main() 35 { 36 vector<int>::iterator pos; 37 int m,k,x; 38 scanf("%d%d",&m,&k); 39 while (m--) 40 { 41 scanf("%s%d",mode,&x); 42 if (strcmp(mode,"add")==0) 43 { 44 //lower_bound*(>=)相当于二分，相当方便！！！ upper_bound(>) 45 pos=lower_bound(st.begin(),st.end(),x); 46 if ((pos>=st.end() || *pos-x>k) && 47 ((st.begin()==st.end() || pos-1<st.begin()) || x - *(pos-1)>k)) 48 st.insert(pos,x); 49 } 50 else if (strcmp(mode,"del")==0) 51 { 52 pos=lower_bound(st.begin(),st.end(),x); 53 if (!(pos>=st.end() || *pos-x>k)) 54 st.erase(pos); 55 if (!((st.begin()==st.end() || pos-1<st.begin()) || x - *(pos-1)>k)) 56 st.erase(pos-1); 57 } 58 else 59 { 60 pos=lower_bound(st.begin(),st.end(),x); 61 if ((pos>=st.end() || *pos-x>k) && 62 ((st.begin()==st.end() || pos-1<st.begin()) || x - *(pos-1)>k)) 63 printf("No\n"); 64 else 65 printf("Yes\n"); 66 } 67 } 68 return 0; 69 } 70 /* 71 14 1 72 add 1 73 add 3 74 add 5 75 add 7 76 add 9 77 query 2 78 del 2 79 query 2 80 query 8 81 del 8 82 query 8 83 query 6 84 del 5 85 query 5 86 */
H.
最小生成树裸题
1 /* 2 最小生成树裸题 3 边少，用Kruskal 4 */ 5 #include <cstdio> 6 #include <cstdlib> 7 #include <cmath> 8 #include <cstring> 9 #include <time.h> 10 #include <string> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <stack> 15 #include <queue> 16 #include <vector> 17 #include <bitset> 18 #include <ext/rope> 19 #include <algorithm> 20 #include <iostream> 21 using namespace std; 22 #define ll long long 23 #define minv 1e-6 24 #define inf 1e9 25 #define pi 3.1415926536 26 #define E 2.7182818284 27 const ll mod=1e9+7;//998244353 28 const int maxn=1e5+10; 29 30 struct node 31 { 32 int a,b,v; 33 }e[maxn*5];//注意 34 int fa[maxn]; 35 36 int getfather(int d) 37 { 38 if (fa[d]==d) 39 return d; 40 fa[d]=getfather(fa[d]); 41 return fa[d]; 42 } 43 44 int cmp(node a,node b) 45 { 46 return a.v<b.v; 47 } 48 49 int g=0,b[maxn]; 50 51 int main() 52 { 53 int n,m,i,c=0,x,y,sum=0; 54 scanf("%d%d",&n,&m); 55 for (i=1;i<=n;i++) 56 fa[i]=i; 57 for (i=1;i<=m;i++) 58 scanf("%d%d%d",&e[i].a,&e[i].b,&e[i].v); 59 sort(e+1,e+m+1,cmp); 60 c=n-1; 61 i=0; 62 while (c--) 63 { 64 do 65 { 66 i++; 67 x=getfather(e[i].a); 68 y=getfather(e[i].b); 69 } 70 while (x==y); 71 fa[x]=y; 72 sum+=e[i].v; 73 g++; 74 b[g]=e[i].v; 75 } 76 printf("%d",sum); 77 78 sort(b+1,b+g+1); 79 printf("\ng=%d\n",g); 80 for (i=1;i<=g;i++) 81 printf("%d\n",b[i]); 82 return 0; 83 } 84 /* 85 4 5 86 1 2 3 87 1 3 1 88 1 2 3 89 2 3 4 90 1 4 5 91 */
1 /* 2 最小生成树裸题 3 Prim + 堆优化 4 */ 5 #include <cstdio> 6 #include <cstdlib> 7 #include <cmath> 8 #include <cstring> 9 #include <time.h> 10 #include <string> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <stack> 15 #include <queue> 16 #include <vector> 17 #include <bitset> 18 #include <ext/rope> 19 #include <algorithm> 20 #include <iostream> 21 using namespace std; 22 #define ll long long 23 #define minv 1e-6 24 #define inf 1e9 25 #define pi 3.1415926536 26 #define E 2.7182818284 27 const ll mod=1e9+7;//998244353 28 const int maxn=1e5+10; 29 30 int dist[maxn]; 31 struct rec 32 { 33 int d,dist; 34 }; 35 36 struct cmp1 37 { 38 bool operator() (rec a,rec b) 39 { 40 return a.dist>b.dist; //最小堆 > 而不是 < 41 } 42 }; 43 44 ///注意，若只用d,不用dist，结果不对！！！ 45 priority_queue<rec,vector<rec>,cmp1 >st; 46 47 struct node 48 { 49 int d,len; 50 node* next; 51 }*e[maxn],*p; 52 bool vis[maxn]={0}; 53 54 int g=0,b[maxn]; 55 56 int main() 57 { 58 int n,m,i,d,x,y,z,sum=0; 59 vector<pair<int,int> >::iterator j; 60 scanf("%d%d",&n,&m); 61 for (i=1;i<=n;i++) 62 e[i]=NULL; 63 for (i=1;i<=m;i++) 64 { 65 scanf("%d%d%d",&x,&y,&z); 66 p=(node*) malloc (sizeof(node)); 67 p->d=y; 68 p->len=z; 69 p->next=e[x]; 70 e[x]=p; 71 72 p=(node*) malloc (sizeof(node)); 73 p->d=x; 74 p->len=z; 75 p->next=e[y]; 76 e[y]=p; 77 } 78 for (i=1;i<=n;i++) 79 dist[i]=inf; 80 dist[1]=0; 81 st.push({1,0}); 82 while (1) 83 { 84 while (!st.empty() && vis[st.top().d]) 85 st.pop(); 86 if (st.empty()) 87 break; 88 d=st.top().d; 89 st.pop(); 90 vis[d]=1; 91 sum+=dist[d]; 92 p=e[d]; 93 while (p) 94 { 95 if (dist[p->d]>p->len) 96 { 97 dist[p->d]=p->len; 98 st.push({p->d,p->len}); 99 } 100 p=p->next; 101 } 102 } 103 printf("%d",sum); 104 return 0; 105 }
J.
最短路径裸题
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <time.h> 6 #include <string> 7 #include <set> 8 #include <map> 9 #include <list> 10 #include <stack> 11 #include <queue> 12 #include <vector> 13 #include <bitset> 14 #include <ext/rope> 15 #include <algorithm> 16 #include <iostream> 17 using namespace std; 18 #define ll long long 19 #define minv 1e-6 20 #define inf 1e9 21 #define pi 3.1415926536 22 #define E 2.7182818284 23 const ll mod=1e9+7;//998244353 24 const int maxn=1e5+10; 25 26 vector<pair<int,int> >e[maxn]; 27 int dist[maxn],q[maxn]; 28 bool vis[maxn]={0}; 29 30 int main() 31 { 32 int n,m,d,t,i,x,y,z,head,tail; 33 vector<pair<int,int> >::iterator j; 34 scanf("%d%d%d%d",&n,&m,&d,&t); 35 for (i=1;i<=m;i++) 36 { 37 scanf("%d%d%d",&x,&y,&z); 38 e[x].push_back(make_pair(y,z)); 39 e[y].push_back(make_pair(x,z)); 40 } 41 for (i=1;i<=n;i++) 42 dist[i]=inf; 43 dist[d]=0; 44 //я╜╩╥╤сап 45 head=0,tail=1; 46 q[1]=d; 47 while (head!=tail) 48 { 49 head=(head+1)%n; 50 d=q[head]; 51 for (j=e[d].begin();j!=e[d].end();j++) 52 if (dist[j->first]>dist[d]+j->second) 53 { 54 dist[j->first]=dist[d]+j->second; 55 if (!vis[j->first]) 56 { 57 vis[j->first]=1; 58 tail=(tail+1)%n; 59 q[tail]=j->first; 60 } 61 } 62 vis[d]=0; 63 } 64 if (dist[t]==inf) 65 printf("-1"); 66 else 67 printf("%d",dist[t]); 68 return 0; 69 }
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <time.h> 6 #include <string> 7 #include <set> 8 #include <map> 9 #include <list> 10 #include <stack> 11 #include <queue> 12 #include <vector> 13 #include <bitset> 14 #include <ext/rope> 15 #include <algorithm> 16 #include <iostream> 17 using namespace std; 18 #define ll long long 19 #define minv 1e-6 20 #define inf 1e9 21 #define pi 3.1415926536 22 #define E 2.7182818284 23 const ll mod=1e9+7;//998244353 24 const int maxn=1e5+10; 25 26 struct rec 27 { 28 int d,dist; 29 }; 30 31 struct cmp 32 { 33 bool operator() (rec a,rec b) 34 { 35 return a.dist>b.dist; 36 } 37 }; 38 39 priority_queue<rec,vector<rec>,cmp>st; 40 41 struct node 42 { 43 int d,len; 44 node* next; 45 }*e[maxn],*p; 46 bool vis[maxn]={0}; 47 int dist[maxn]; 48 49 int main() 50 { 51 int n,m,d,t,i,x,y,z; 52 scanf("%d%d%d%d",&n,&m,&d,&t); 53 for (i=1;i<=n;i++) 54 e[i]=NULL; 55 for (i=1;i<=m;i++) 56 { 57 scanf("%d%d%d",&x,&y,&z); 58 p=(node*) malloc (sizeof(node)); 59 p->d=y; 60 p->len=z; 61 p->next=e[x]; 62 e[x]=p; 63 64 p=(node*) malloc (sizeof(node)); 65 p->d=x; 66 p->len=z; 67 p->next=e[y]; 68 e[y]=p; 69 } 70 for (i=1;i<=n;i++) 71 dist[i]=inf; 72 dist[d]=0; 73 st.push({d,0}); 74 while (d!=t) 75 { 76 while (!st.empty() && vis[st.top().d]) 77 st.pop(); 78 if (st.empty()) 79 break; 80 d=st.top().d; 81 st.pop(); 82 p=e[d]; 83 while (p) 84 { 85 if (dist[p->d]>dist[d]+p->len) 86 { 87 dist[p->d]=dist[d]+p->len; 88 st.push({p->d,dist[p->d]}); 89 } 90 p=p->next; 91 } 92 } 93 if (dist[t]==inf) 94 printf("-1"); 95 else 96 printf("%d",dist[t]); 97 return 0; 98 }
J.
可用矩阵快速幂
1 /* 2 t(i)->t(i+1) 一次多元函数 矩阵快速幂是套路 3 4 n个变量和常数，则创建(n+1)*(n+1)的矩阵 5 t(i+1)=t(i)+f(i) 6 f(i+1)=f(i)*k+p 7 见目录图片 8 */ 9 #include <cstdio> 10 #include <cstdlib> 11 #include <cmath> 12 #include <cstring> 13 #include <time.h> 14 #include <string> 15 #include <set> 16 #include <map> 17 #include <list> 18 #include <stack> 19 #include <queue> 20 #include <vector> 21 #include <bitset> 22 #include <ext/rope> 23 #include <algorithm> 24 #include <iostream> 25 using namespace std; 26 #define ll long long 27 #define minv 1e-6 28 #define inf 1e9 29 #define pi 3.1315926536 30 #define E 2.7182818283 31 const ll mod=1e9+7;//998233353 32 const int maxn=1e5+10; 33 34 //要加{} 35 ll a[3][3]={ 36 {1,0,0}, 37 {0,0,0}, 38 {0,0,1} 39 }; 40 ll b[3][3],c[3][3]; 41 42 int main() 43 { 44 int n,i,j,k; 45 ll K,p; 46 scanf("%d%lld%lld",&n,&K,&p); 47 a[0][1]=a[1][1]=K; 48 a[0][2]=a[1][2]=p; 49 for (i=0;i<3;i++) 50 for (j=0;j<3;j++) 51 b[i][j]=(i==j); 52 n--; 53 while (n) 54 { 55 if (n & 1) 56 { 57 for (i=0;i<3;i++) 58 for (j=0;j<3;j++) 59 c[i][j]=b[i][j],b[i][j]=0; 60 for (i=0;i<3;i++) 61 for (j=0;j<3;j++) 62 for (k=0;k<3;k++) 63 b[i][j]=(b[i][j]+c[i][k]*a[k][j])%mod; 64 } 65 66 for (i=0;i<3;i++) 67 for (j=0;j<3;j++) 68 c[i][j]=a[i][j],a[i][j]=0; 69 for (i=0;i<3;i++) 70 for (j=0;j<3;j++) 71 for (k=0;k<3;k++) 72 a[i][j]=(a[i][j]+c[i][k]*c[k][j])%mod; 73 n>>=1; 74 } 75 printf("%lld",(b[0][0]+b[0][1]+b[0][2])%mod); 76 return 0; 77 }
