A 第一个水题
要知道units digit的意思 （个位数）
有一点点小繁琐。
1 #include <iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<stdlib.h>
6 #include<cmath>
7 using namespace std;
8 #define N 35
9 int a[N],b[N];
10 int o1[N],o2[N];
11 struct node
12 {
13 int di;
14 int v;
15 }p[N];
16 int judge(int x,int y,int n)
17 {
18 int i;
19 for(i = 0 ;i < n ; i++)
20 {
21 int g = 0;
22 while(x+g<=y&&o2[i+g]==o1[x+g])
23 g++;
24 // if(x==1&&y==2)
25 // printf("%d, %d %d %d\n ",i,g,o2[i],o1[x]);
26 if(x+g>y)
27 return 1;
28 }
29 return 0;
30 }
31 bool cmp(node a,node b)
32 {
33 if(a.di==b.di)
34 return a.v<b.v;
35 return a.di<b.di;
36 }
37 bool ccmp(int a,int b)
38 {
39 return a>b;
40 }
41 int main()
42 {
43 int t,i,j,n1,n2;
44 cin>>t;
45 while(t--)
46 {
47 scanf("%d%d",&n1,&n2);
48 for(i = 0;i < n1 ;i++)
49 scanf("%d",&a[i]);
50 for(i = 0 ;i< n2 ; i++)
51 scanf("%d",&b[i]);
52 sort(a,a+n1,ccmp); sort(b,b+n2,ccmp);
53 int g1=1,g2=1,g;
54 o1[0] = a[0];
55 for(i = 1; i < n1 ; i++)
56 if(a[i]!=a[i-1])
57 {
58 o1[g1++] = a[i];
59 }
60 o2[0] = b[0];
61 for(i = 1;i < n2 ; i++)
62 if(b[i]!=b[i-1])
63 {
64 o2[g2++] = b[i];
65 }
66 int st=-1,en=-1,mz=-1,maxz=-1;
67 for(i = 0; i < g1 ; i++)
68 {
69 for(j = i ; j < g1 ; j++)
70 {
71 if(judge(i,j,g2))
72 {
73 // cout<<i<<" "<<j<<endl;
74 int mm = -1;
75 for(int e = i ; e <= j; e++)
76 mm = max(mm,o1[e]);
77 if(maxz<=j-i+1)
78 {
79 if(maxz<j-i+1||mm>mz)
80 {
81 st = i;
82 en = j;
83 mz = mm;
84 maxz = j-i+1;
85 }
86 }
87 }
88 }
89 }
90 if(st==-1)
91 {
92 puts("NONE");
93 continue;
94 }
95 g = 0;
96 for(i = st ; i<= en; i++)
97 {
98 p[g].v = o1[i];
99 p[g].di = o1[i]%10;
100 g++;
101 }
102 for(i = 0 ;i < g-1 ; i++)
103 printf("%d ",p[i].v);
104 printf("%d\n",p[i].v);
105 sort(p,p+g,cmp);
106 for(i = 0 ;i < g-1 ; i++)
107 printf("%d ",p[i].v);
108 printf("%d\n",p[i].v);
109 }
110 return 0;
111 }
B
队友过的，待填。。
C
卡时限的奇葩题。
预处理出以k点为最高价格点时到其它点的单源最短路径，在询问时用这个n种情况依次更新u,v；
minz = min(minz,w[k][u]+w[k][v]+p[k]);
直接写可能会超时，反正我超了N久，需要把询问先读进来，再依次更新。
1 #include <iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<stdlib.h>
6 #include<vector>
7 #include<cmath>
8 #include<queue>
9 #include<set>
10 using namespace std;
11 #define N 1010
12 #define M 20010
13 #define LL long long
14 #define INF 1e18
15 const double eps = 1e-8;
16 const double pi = acos(-1.0);
17 const double inf = ~0u>>2;
18 vector<int>ed[N][2];
19 int cot[N],co[N][N];
20 LL w[N][N],dis[N],mm[M];
21 bool vis[N];
22 struct node
23 {
24 int c,id;
25 }p[N];
26 struct nodd
27 {
28 int u,v;
29 }qi[M];
30 struct mode
31 {
32 int u;
33 LL w;
34 friend bool operator <(const mode &a,const mode &b)
35 {
36 return a.w>b.w;
37 }
38 };
39 void spfa(int st,int n)
40 {
41 memset(vis,0,sizeof(vis));
42 queue<mode>q;
43 int i;
44 mode tt,ts;
45 for(i = 1; i <= n; i++)
46 {
47 dis[i] = INF;
48 }
49 ts.u = st;
50 ts.w = 0;
51 q.push(ts);
52 vis[st] = 1;
53 dis[st] = 0;
54 while(!q.empty())
55 {
56 tt = q.front();
57 int u = tt.u;
58 q.pop();
59 vis[u] = 0;
60 for(i = 0 ;i < ed[u][0].size(); i++)
61 {
62 int v = ed[u][0][i];
63 int w = ed[u][1][i];
64 if(p[v].c>p[st].c) continue;
65 if(dis[v]>dis[u]+w)
66 {
67 dis[v] = dis[u]+w;
68 if(!vis[v])
69 {
70 ts.u = v;
71 ts.w = dis[v];
72 vis[v] = 1;
73 q.push(ts);
74 }
75 }
76 }
77 }
78 }
79 bool cmp(node a,node b)
80 {
81 return a.c>b.c;
82 }
83 int main()
84 {
85 int n,m,i,j;
86 while(scanf("%d%d",&n,&m)&&n&&m)
87 {
88 for(i = 1; i <= n ; i++)
89 {
90 scanf("%d",&p[i].c);
91 p[i].id = i;
92 ed[i][0].clear();
93 ed[i][1].clear();
94 }
95 for(i = 1; i <= m; i++)
96 {
97 int u,v,w;
98 scanf("%d%d%d",&u,&v,&w);
99 ed[u][0].push_back(v);
100 ed[u][1].push_back(w);
101 ed[v][0].push_back(u);
102 ed[v][1].push_back(w);
103 }
104 ;
105 int q;
106 scanf("%d",&q);
107 for(i = 1; i <= q ; i++)
108 {
109 scanf("%d%d",&qi[i].u,&qi[i].v);
110 mm[i] = INF;
111 }
112 for(i = 1 ;i <= n ; i++)
113 {
114 spfa(i,n);
115 for(j = 1; j <= q ; j++)
116 {
117 mm[j] = min(mm[j],dis[qi[j].u]+dis[qi[j].v]+p[i].c);
118 }
119 }
120 for(i = 1; i < q; i++)
121 {
122 if(mm[i]==INF)
123 printf("-1\n");
124 else
125 printf("%lld\n",mm[i]);
126 }
127 if(mm[q]==INF)
128 puts("-1");
129 else
130 printf("%lld\n",mm[q]);
131 puts("");
132 }
133 return 0;
134 }
D
线段树水题，有段时间没写，居然lz忘记更新，卡了几个小时。。。
1 #include <iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<stdlib.h>
6 #include<cmath>
7 using namespace std;
8 #define N 200010
9 #define INF 0xfffffff
10 int a[N],b[N],lz[N<<2];
11 int s[N<<2];
12 void up(int w)
13 {
14 s[w] = max(s[w<<1],s[w<<1|1]);
15 }
16 void build(int l,int r,int w)
17 {
18 lz[w] = 0;
19 if(l==r)
20 {
21 s[w] = a[l];
22 return ;
23 }
24 int m = (l+r)>>1;
25 build(l,m,w<<1);
26 build(m+1,r,w<<1|1);
27 up(w);
28 }
29 void down(int w,int m)
30 {
31 if(lz[w]!=0)
32 {
33 s[w<<1]+=lz[w];
34 s[w<<1|1]+=lz[w];
35 lz[w<<1]+=lz[w];
36 lz[w<<1|1]+=lz[w];
37 lz[w] = 0;
38 }
39 }
40 void update(int x,int y,int d,int l,int r,int w)
41 {
42 if(x<=l&&y>=r)
43 {
44 s[w] += d;
45 lz[w] += d;
46 return ;
47 }
48 down(w,r-l+1);
49 int m = (l+r)>>1;
50 if(x<=m)
51 update(x,y,d,l,m,w<<1);
52 if(y>m)
53 update(x,y,d,m+1,r,w<<1|1);
54 up(w);
55 }
56 int query(int x,int y,int l,int r,int w)
57 {
58 if(x<=l&&y>=r)
59 return s[w];
60 int m =(l+r)>>1;
61 down(w,r-l+1);
62 int res = -INF;
63 if(x<=m)
64 res = max(query(x,y,l,m,w<<1),res);
65 if(y>m)
66 res = max(query(x,y,m+1,r,w<<1|1),res);
67 return res;
68 }
69 int main()
70 {
71 int i,n,t,m,k;
72 scanf("%d",&t);
73 while(t--)
74 {
75 scanf("%d%d%d",&n,&m,&k);
76 for(i = 1 ;i <= n; i++)
77 scanf("%d",&b[i]);
78 a[n+1] = 0;
79 int g = n;
80 for(i = n ; i >= 1 ; i--)
81 {
82 a[i] = a[i+1]+b[i];
83 if(n-i+1>k)
84 {
85 a[i]-=b[g--];
86 }
87 }
88 n = n-k+1;
89 build(1,n,1);
90 while(m--)
91 {
92 int x,y,z;
93 scanf("%d%d%d",&z,&x,&y);
94 if(z==0)
95 {
96 update(max(x-k+1,1),min(x,n),y-b[x],1,n,1);
97 b[x] = y;
98 }
99 else if(z==1)
100 {
101 update(max(x-k+1,1),min(x,n),b[y]-b[x],1,n,1);
102 update(max(y-k+1,1),min(y,n),b[x]-b[y],1,n,1);
103 swap(b[x],b[y]);
104 }
105 else
106 printf("%d\n",query(x,y-k+1,1,n,1));
107 }
108 }
109 return 0;
110 }
E
简单几何题，枚举所在点到所有线段端点的直线，可以想到最多穿墙的线肯定是过端点的。
注意一点，给出的线段会有与你枚举的直线重合的线段
1 #include <iostream>
2 #include <algorithm>
3 #include <cmath>
4 #include <cstring>
5 #include <string>
6 #include <cstdio>
7 using namespace std;
8 #define eps 1e-8
9 #define zero(x) (((x)>0?(x):-(x)) < eps)
10 struct point
11 {
12 double x,y;
13 }p[4001];
14 struct line
15 {
16 point a,b;
17 };
18 double dis(point p1,point p2)
19 {
20 return (p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y);
21 }
22 double xmult(point p1,point p2,point p0)
23 {
24 return (p1.x-p0.x)*(p2.y-p0.y) - (p2.x-p0.x)*(p1.y-p0.y);
25 }
26 int pinline(point p,line l)
27 {
28 return zero(xmult(p,l.a,l.b))&&(l.a.x-p.x)*(l.b.x-p.x) < eps&&(l.a.y-p.y)*(l.b.y-p.y) < eps;
29 }
30 point inter(point u1,point u2,point v1,point v2)
31 {
32 point ret = u1;
33 double t = ((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))
34 /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
35 ret.x += (u2.x-u1.x)*t;
36 ret.y += (u2.y-u1.y)*t;
37 return ret;
38 }
39 int pingxing(line u,line v)
40 {
41 return zero((u.a.x-u.b.x)*(v.a.y-v.b.y)-(v.a.x-v.b.x)*(u.a.y-u.b.y));
42 }
43 int nopoint(point p,line l)
44 {
45 return pinline(p,l)&&(!zero(p.x-l.a.x)||!zero(p.y-l.a.y))&&(!zero(p.x-l.b.x)||!zero(p.y-l.b.y));
46 }
47 int main()
48 {
49 int n,t,i,j,ans,num;
50 scanf("%d",&t);
51 while(t--)
52 {
53 scanf("%d",&n);
54 for(i = 1;i <= 2*n;i ++)
55 {
56 scanf("%lf%lf",&p[i].x,&p[i].y);
57 }
58 ans = 0;
59 scanf("%lf%lf",&p[0].x,&p[0].y);
60 for(i = 1;i <= 2*n;i ++)
61 {
62 num = 0;
63 for(j = 1;j <= n;j ++)
64 {
65 line u,v,ti;
66 point temp;
67 u.a = p[0];
68 u.b = p[i];
69 v.a = p[2*j-1];
70 v.b = p[2*j];
71 ti.a = p[0];
72 ti.b = p[i];
73 if(pinline(p[2*j],ti))
74 {
75 num++;
76 continue;
77 }
78 ti.b = p[2*j];
79 if(pinline(p[i],ti))
80 {
81 num++;
82 continue;
83 }
84 if(pingxing(u,v)) continue;
85 temp = inter(p[0],p[i],p[2*j-1],p[2*j]);
86 if(pinline(temp,v))
87 {
88 line tt;
89 tt.a = p[0];
90 tt.b = temp;
91 if(pinline(p[i],tt))
92 num ++;
93 else
94 {
95 tt.a = p[0];
96 tt.b = p[i];
97 if(pinline(temp,tt))
98 num++;
99 }
100 }
101 }
102 ans = max(ans,num);
103 }
104 printf("%d\n",ans);
105 }
106 return 0;
107 }
H
ac自动机，比赛的时候没来得及看，看了也不一定会。。
这个题不管 去除子串这个条件就是裸自动机，可以先进行裸自动机把需要的串标记出来，再在已经标记的串里面进行自动机的查找去掉它所含有的子串。
因为数据挺大，可以省掉strlen()这个操作，全改为指针操作。
模板里面有句话错了，wa了N久。。
1 #include <iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<stdlib.h>
6 #include<vector>
7 #include<cmath>
8 #include<queue>
9 #include<set>
10 using namespace std;
11 #define N 3100010
12 #define M 1100
13 #define NN 5100000
14 #define LL long long
15 #define INF 0xfffffff
16 const double eps = 1e-8;
17 const double pi = acos(-1.0);
18 const double inf = ~0u>>2;
19 const int child_num = 26;
20 bool o[2510];
21 char s[NN],sr[NN],vir[M];
22 char dig[20],ss[2510][M];
23 class ACAutomo
24 {
25 private:
26 int ch[N][child_num];
27 int val[N];
28 int fail[N];
29 int Q[N];
30 int id[128];
31 int sz;
32 public:
33 void init()
34 {
35 fail[0] = 0;
36 for(int i = ‘A‘; i <= ‘Z‘ ; i++)
37 id[i] = i-‘A‘;
38 }
39 void reset()
40 {
41 memset(ch[0],0,sizeof(ch[0]));
42 memset(o,0,sizeof(o));
43 sz = 1;
44 }
45 void insert(char *a,int key)
46 {
47 int p = 0 ;
48 for( ; *a !=‘\0‘; a++)
49 {
50 int d = id[*a];
51 if(ch[p][d]==0)
52 {
53 memset(ch[sz],0,sizeof(ch[sz]));
54 val[sz] = 0;
55 ch[p][d] = sz++;
56 }
57 p = ch[p][d];
58 }
59 val[p]=key;
60 }
61 void construct()
62 {
63 int i,head=0,tail = 0;
64 for(i = 0 ; i < child_num ; i++)
65 {
66 if(ch[0][i])
67 {
68 Q[tail++] = ch[0][i];
69 fail[ch[0][i]] = 0;
70 }
71 }
72 while(head!=tail)
73 {
74 int u = Q[head++];
75 for(i = 0; i < child_num ; i++)
76 {
77 if(ch[u][i])
78 {
79 Q[tail++] = ch[u][i];
80 fail[ch[u][i]] = ch[fail[u]][i];
81 }
82 else
83 ch[u][i] = ch[fail[u]][i];
84 }
85 }
86 }
87 void work(char *s)
88 {
89 int p = 0;
90 for( ; *s!=‘\0‘ ; s++)
91 {
92 int d = id[*s];
93 p = ch[p][d];
94 int tmp = p;
95 if(val[tmp])
96 {
97 o[val[tmp]] = 1;
98 continue;
99 }
100 while(tmp!=0&&val[tmp]!=0)
101 {
102 o[val[tmp]] = 1;
103 tmp = fail[tmp];
104 }
105 }
106 }
107 void solve(char *s)
108 {
109 int p = 0;
110 for( ; *s!=‘\0‘ ; s++)
111 {
112 int d = id[*s];
113 p = ch[p][d];
114 int tmp = p;
115 while(tmp!=0&&val[tmp]!=0)
116 {
117 o[val[tmp]] = 0;
118 tmp = fail[tmp];
119 }
120 }
121 }
122 } ac;
123 void change(char *sr,int k)
124 {
125 int g=0;
126 for( ; *sr!=‘\0‘ ; sr++)
127 {
128 if((*sr)>=‘A‘&&(*sr)<=‘Z‘)
129 {
130 if(k!=-1)
131 ss[k][g++] = (*sr);
132 else
133 s[g++] = (*sr);
134 }
135 else if((*sr)==‘[‘)
136 {
137 sr++;
138 int e = 0;
139 while((*sr)>=‘0‘&&(*sr)<=‘9‘)
140 {
141 dig[e++] = (*sr);
142 sr++;
143 }
144 dig[e] = ‘\0‘;
145 int num = atoi(dig);
146 while(num--)
147 {
148 if(k!=-1)
149 ss[k][g++] = (*sr);
150 else
151 s[g++] = (*sr);
152 }
153 sr++;
154 }
155 }
156 if(k!=-1)
157 ss[k][g] = ‘\0‘;
158 else
159 s[g] = ‘\0‘;
160 }
161 int main()
162 {
163 int n,i;
164 ac.init();
165 int t;
166 scanf("%d",&t);
167 int tt = 0;
168 while(t--)
169 {
170 tt++;
171 scanf("%d",&n);
172 ac.reset();
173 for(i = 1; i <= n; i++)
174 {
175 scanf("%s",vir);
176 change(vir,i);
177 ac.insert(ss[i],i);
178 }
179 //cout<<n<<endl;
180 ac.construct();
181 scanf("%s",sr);
182 // if(tt>12)
183 // {
184 // puts("1");
185 // continue;
186 // }
187 change(sr,-1);
188 ac.work(s);
189 for(i = 1 ; i <= n ; i++)
190 {
191 if(o[i])
192 {
193 ac.solve(ss[i]);
194 o[i] = 1;
195 }
196 }
197 int num = 0;
198 for(i = 1; i <= n; i++)
199 if(o[i]) num++;
200 printf("%d\n",num);
201 }
202 return 0;
203 }
2012金华邀请赛
时间： 2024-10-05 04:19:16