2017 多校联合训练 6 题解
Problem 1001
对于每个询问，根据后缀{前缀插入到AC自动机内
然后每个单词写2遍，中间用{隔开，统计有那几个前缀和后缀
最后对于每个前缀和后缀 输出方案数
注意：单词可能有重复，可以用vector记录一下end数组
1 #include<iostream> 2 #include<algorithm> 3 #include<cmath> 4 #include<cstring> 5 #include<queue> 6 #include<string> 7 #define next nxt 8 #define end ed 9 using namespace std; 10 int next[1200001][27],fail[1200001],dep[1200001]; 11 vector<int> end[1200001]; 12 int root,len; 13 int newnode(){ 14 for(int i=0;i<27;i++) 15 next[len][i]=-1; 16 end[len].clear(); 17 len++; 18 return len-1; 19 } 20 void clear(){ 21 len=0; 22 root=newnode(); 23 return; 24 } 25 void insert(char s[],int num){ 26 int ls=strlen(s); 27 int now=root; 28 for(int i=0;i<ls;i++){ 29 if(next[now][s[i]-'a']==-1) 30 next[now][s[i]-'a']=newnode(); 31 now=next[now][s[i]-'a']; 32 dep[now]=i+1; 33 } 34 end[now].push_back(num); 35 } 36 void build(){ 37 queue<int>q; 38 int i; 39 fail[root]=root; 40 for(i=0;i<27;i++){ 41 if(next[root][i]==-1) 42 next[root][i]=root; 43 else{ 44 fail[next[root][i]]=root; 45 q.push(next[root][i]); 46 } 47 } 48 while(!q.empty()){ 49 int now=q.front(); 50 q.pop(); 51 for(i=0;i<27;i++){ 52 if(next[now][i]==-1) 53 next[now][i]=next[fail[now]][i]; 54 else{ 55 fail[next[now][i]]=next[fail[now]][i]; 56 q.push(next[now][i]); 57 } 58 } 59 } 60 return; 61 } 62 int vis[1000010]; 63 void ask(char c[],int n,int num){ 64 bool flag1=false; 65 int lc=num; 66 int now=root; 67 int i; 68 for(i=0;i<lc;i++){ 69 now=next[now][c[i]-'a']; 70 int temp=now; 71 while(temp!=root){ 72 if(end[temp].size()&&dep[temp]<=n){ 73 flag1=1; 74 for (auto u:end[temp]) 75 vis[u]++; 76 } 77 temp=fail[temp]; 78 } 79 } 80 } 81 char s[2000001],t[2000001],ss[2000001],tt[2000001]; 82 char *si[100010]; 83 int n,m,le[100010],le2[100010]; 84 int main(){ 85 int i,j; 86 int _; 87 scanf("%d",&_); 88 while(_--){ 89 scanf("%d%d",&n,&m); 90 clear(); 91 for (i=1,j=0;i<=n;i++) 92 { 93 si[i]=ss+j; 94 scanf("%s",si[i]); 95 le[i]=strlen(si[i])+1; 96 j+=le[i]; 97 strcpy(ss+j,si[i]); 98 ss[j-1]='z'+1; 99 j+=le[i]; 100 101 le2[i]=strlen(si[i]); 102 } 103 for(i=1;i<=m;i++) 104 { 105 s[0]='z'+1; 106 scanf("%s%s",s+1,t); 107 strcat(t,s); 108 insert(t,i); 109 } 110 build(); 111 memset(vis,0,sizeof(vis)); 112 int ans=0; 113 //for(i=1;i<=m;i++){ 114 for (i=1;i<=n;i++) 115 { 116 //strcpy(tt,si[i].c_str()); 117 ask(si[i],le[i],le2[i]); 118 } 119 //} 120 for (i=1;i<=m;i++) 121 printf("%d\n",vis[i]); 122 } 123 return 0; 124 }
Problem 1002
这里不总是在中垂线上的点取到最小值
通过观察可发现，当两个点离半径的距离到一定的程度时答案保持不变
算出这个距离即可
另外要注意这两个点可能会重合
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cstring> 7 #include<string> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #include<queue> 12 using namespace std; 13 const double eps=1e-7; 14 #define y1 jkhjk 15 #define y2 erjkf 16 #define y3 jkdfhdw 17 #define y4 jkhjkff 18 #define y5 kjhkk 19 int _; 20 double r,x1,y1,x2,y2; 21 int main() 22 { 23 scanf("%d",&_); 24 while (_--) 25 { 26 scanf("%lf",&r); 27 scanf("%lf%lf",&x1,&y1); 28 scanf("%lf%lf",&x2,&y2); 29 if (x1==0&&y1==0) 30 { 31 printf("%.9f\n",r*2); 32 continue; 33 } 34 if (fabs(x1-x2)<=eps&&fabs(y1-y2)<=eps) 35 { 36 double rest=r-sqrt(x1*x1+y1*y1); 37 printf("%.9f\n",rest*2); 38 continue; 39 } 40 double x3=(x1+x2)/2; 41 double y3=(y1+y2)/2; 42 if (fabs(y1-y2)<=eps) 43 { 44 double x4=x3; 45 double y4=sqrt(r*r-x3*x3); 46 double x5=x3; 47 double y5=-sqrt(r*r-x3*x3); 48 double x6,y6,x7,y7; 49 double bili=sqrt(x1*x1+y1*y1)/r; 50 x6=x1/bili; 51 y6=y1/bili; 52 x7=x2/bili; 53 y7=y2/bili; 54 double dist=sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)); 55 double pp=sqrt(x1*x1+y1*y1); 56 double ans=9999999; 57 double sita=(dist/(2*pp)); 58 //cout<<sita<<endl; 59 sita=sqrt(1.0-sita*sita); 60 if (sqrt(x1*x1+y1*y1)>=r*sita-eps) 61 ans=sqrt((x7-x6)*(x7-x6)+(y7-y6)*(y7-y6)); 62 ans=min(ans,sqrt((x4-x1)*(x4-x1)+(y4-y1)*(y4-y1))*2); 63 ans=min(ans,sqrt((x5-x1)*(x5-x1)+(y5-y1)*(y5-y1))*2); 64 printf("%.9f\n",ans); 65 66 } 67 else 68 { 69 double k; 70 if (x1==x2) k=0; 71 else 72 k=(x2-x1)/(y1-y2); 73 double a=1+k*k; 74 double b=2*k*y3-2*k*k*x3; 75 double c=k*k*x3*x3-2*k*x3*y3+y3*y3-r*r; 76 double det=b*b-4*a*c; 77 double x4=(-b+sqrt(det))/(2*a); 78 double y4=k*(x4-x3)+y3; 79 double x5=(-b-sqrt(det))/(2*a); 80 double y5=k*(x5-x3)+y3; 81 double x6,y6,x7,y7; 82 double bili=sqrt(x1*x1+y1*y1)/r; 83 x6=x1/bili; 84 y6=y1/bili; 85 x7=x2/bili; 86 y7=y2/bili; 87 double dist=sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)); 88 double pp=sqrt(x1*x1+y1*y1); 89 double ans=9999999; 90 double sita=(dist/(2*pp)); 91 sita=sqrt(1.0-sita*sita); 92 if (sqrt(x1*x1+y1*y1)>=r*sita-eps) 93 ans=sqrt((x7-x6)*(x7-x6)+(y7-y6)*(y7-y6)); 94 ans=min(ans,sqrt((x4-x1)*(x4-x1)+(y4-y1)*(y4-y1))*2); 95 ans=min(ans,sqrt((x5-x1)*(x5-x1)+(y5-y1)*(y5-y1))*2); 96 printf("%.9f\n",ans); 97 } 98 } 99 return 0; 100 }
Problem 1003
hnqw1214:
将A数组按照从大到小排序
对于每个下标i
暴力找到最大的不被i整除的数
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cstring> 7 #include<string> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #include<queue> 12 using namespace std; 13 int _,n,a[100010],id[100010]; 14 inline bool cmp(int x,int y) 15 { 16 return a[x]>a[y]; 17 } 18 int main() 19 { 20 scanf("%d",&_); 21 while (_--) 22 { 23 scanf("%d",&n); 24 int i,j; 25 for (i=1;i<=n;i++) 26 { 27 scanf("%d",&a[i]); 28 id[i]=i; 29 } 30 sort(id+1,id+n+1,cmp); 31 for (i=2;i<=n;i++) 32 { 33 int p=0; 34 for (j=1;j<=n;j++) 35 if (id[j]%i) 36 { 37 p=a[id[j]]; 38 break; 39 } 40 if (i<n) printf("%d ",p); 41 else printf("%d\n",p); 42 } 43 } 44 return 0; 45 }
cxhscst2:
预处理出ST表。
一个数的答案被很多区间影响
比如4的答案就是[1, 3], [5, 7], [9, 11]这些区间的最小值中的最小值。
区间个数总和数为nlogn。
ST表分段查询即可。
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)
typedef long long LL;
const int N = 1e5 + 10;
vector <int> v[N];
int cnt = 0;
int f[N][19];
int a[N];
int n;
int ans[N];
int s, t;
int T;
inline int solve(int l, int r){
int k = (int)log2((double)(r - l + 1));
return max(f[l][k], f[r - (1 << k) + 1][k]);
}
void ST(){
rep(i, 1, n) f[i][0] = a[i];
rep(j, 1, 20) rep(i, 1, n)
if ((i + (1 << j) - 1) <= n) f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
}
int main(){
rep(i, 2, 100000){
for (int j = i; j <= 2e5 + 3; j += i) v[i].push_back(j), ++cnt;
}
scanf("%d", &T);
while (T--){
scanf("%d", &n);
rep(i, 1, n) scanf("%d", a + i);
memset(f, 0, sizeof f);
ST();
memset(ans, 0, sizeof ans);
rep(i, 2, n){
s = 1;
for (auto u : v[i]){
t = u - 1;
if (s > n) break;
t = min(t, n);
ans[i] = max(ans[i], solve(s, t));
s = u + 1;
}
}
rep(i, 2, n - 1) printf("%d ", ans[i]); printf("%d\n", ans[n]);
}
return 0;
}
Problem 1006
考虑每个独立块。
Bob只有在每个独立块大小为2的时候才能赢。
所以当n为奇数的时候直接输出Alice
于是这个问题转化成了树上最大匹配问题。
我们求一遍树上最大匹配，结果为tmp。如果tmp <= k + 1则符合题意，Bob赢。
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)
const int N = 1010;
int f[N][2];
vector <int> v[N];
int n, k;
int T;
void dfs(int x){
for (auto u : v[x]){
dfs(u);
int mx = max(f[u][0], f[u][1]);
f[x][0] += mx;
}
for (auto u : v[x]){
int t = f[x][0] - max(f[u][0], f[u][1]) + f[u][0] + 1;
f[x][1] = max(f[x][1], t);
}
}
int main(){
scanf("%d", &T);
while (T--){
scanf("%d%d", &n, &k);
rep(i, 0, n + 1) v[i].clear();
rep(i, 2, n){
int x; scanf("%d", &x);
v[x].push_back(i);
}
if (n & 1){ puts("Alice"); continue; }
memset(f, 0, sizeof f);
dfs(1);
int tmp = max(f[1][0], f[1][1]);
if (tmp * 2 == n && tmp <= k + 1) puts("Bob");
else puts("Alice");
}
return 0;
}
Problem 1008
枚举一下中心向外延伸
如果和超过了一定的值就弹掉中间的位置
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int inf=(1<<30)-1; 4 const int maxn=100010; 5 #define REP(i,n) for(int i=(0);i<(n);i++) 6 #define FOR(i,j,n) for(int i=(j);i<=(n);i++) 7 typedef long long ll; 8 int m; 9 int T; 10 char s[maxn]; 11 int mx=0; 12 13 void cal(int x,int y) 14 { 15 int l=0,r=0,sum=0; 16 while(y-r>x+r){ 17 sum+=abs(s[x+r]-s[y-r]); 18 r++; 19 if(y-l-l-x+1<=2*mx)break; 20 while(sum>m){ 21 sum-=abs(s[x+l]-s[y-l]); 22 l++; 23 } 24 mx=max(mx,r-l); 25 } 26 } 27 int main() 28 { 29 scanf("%d",&T); 30 while(T--) 31 { 32 scanf("%d",&m); 33 scanf("%s",s); 34 mx=0; 35 int n=strlen(s); 36 for(int i=n-1;i>0;i--) cal(0,i); 37 for(int i=1;i<n-1;i++) cal(i,n-1); 38 printf("%d\n",mx); 39 } 40 return 0; 41 }
Problem 1011
直接统计即可
不过要判断一下韦恩图的每个部分都是非负数
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cstring> 7 #include<string> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #include<queue> 12 using namespace std; 13 int _,n; 14 int a[8]; 15 int main() 16 { 17 scanf("%d",&_); 18 while (_--) 19 { 20 scanf("%d",&n); 21 int ans=0; 22 int i,j; 23 for (i=1;i<=n;i++) 24 { 25 int tot=0; 26 bool can=true; 27 for (j=1;j<=7;j++) 28 scanf("%d",&a[j]); 29 tot+=a[7]; 30 if (a[6]>=a[7]) 31 tot+=a[6]-a[7]; 32 else 33 { 34 can=false; 35 continue; 36 } 37 if (a[5]>=a[7]) 38 tot+=a[5]-a[7]; 39 else 40 { 41 can=false; 42 continue; 43 } 44 if (a[4]>=a[7]) 45 tot+=a[4]-a[7]; 46 else 47 { 48 can=false; 49 continue; 50 } 51 if (a[3]>=a[5]+a[6]-a[7]) 52 tot+=a[3]-(a[5]+a[6]-a[7]); 53 else 54 { 55 can=false; 56 continue; 57 } 58 if (a[2]>=a[4]+a[5]-a[7]) 59 tot+=a[2]-(a[4]+a[5]-a[7]); 60 else 61 { 62 can=false; 63 continue; 64 } 65 if (a[1]>=a[4]+a[6]-a[7]) 66 tot+=a[1]-(a[4]+a[6]-a[7]); 67 else 68 { 69 can=false; 70 continue; 71 } 72 if (can) ans=max(ans,tot); 73 } 74 printf("%d\n",ans); 75 } 76 return 0; 77 }
