T1 lcp
题目大意：
q次询问一个字符串两个后缀的最长公共前缀
思路：
可以二分长度hash判断
1 #include<iostream> 2 #include<cmath> 3 #include<algorithm> 4 #include<cstdio> 5 #include<cstring> 6 #include<cstdlib> 7 #include<queue> 8 #include<vector> 9 #include<set> 10 #define MAXN 100100 11 #define ll long long 12 #define inf 2139062143 13 using namespace std; 14 inline int read() 15 { 16 int x=0,f=1;char ch=getchar(); 17 while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();} 18 while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();} 19 return x*f; 20 } 21 const int mod1=19260817,mod2=998244353,bas1=137,bas2=233; 22 ll hsh1[MAXN],hsh2[MAXN],pw1[MAXN],pw2[MAXN]; 23 int n,ans,a,b; 24 char c[MAXN]; 25 int Check(int x) 26 { 27 int va1=(hsh1[a+x-1]-(hsh1[a]*pw1[x-1])%mod1+mod1)%mod1,va2=(hsh2[a+x-1]-(hsh2[a]*pw2[x-1])%mod2+mod2)%mod2; 28 int vb1=(hsh1[b+x-1]-(hsh1[b]*pw1[x-1])%mod1+mod1)%mod1,vb2=(hsh2[b+x-1]-(hsh2[b]*pw2[x-1])%mod2+mod2)%mod2; 29 if(va1==vb1&&va2==vb2) return 1; 30 return 0; 31 } 32 int main() 33 { 34 freopen("lcp.in","r",stdin); 35 freopen("lcp.out","w",stdout); 36 scanf("%s",c+1);n=strlen(c+1); 37 for(int i=pw1[0]=pw2[0]=1;i<=n;i++) 38 pw1[i]=(pw1[i-1]*bas1)%mod1,hsh1[i]=((hsh1[i-1]*bas1)%mod1+c[i]-'a'+1)%mod1, 39 pw2[i]=(pw2[i-1]*bas2)%mod2,hsh2[i]=((hsh2[i-1]*bas2)%mod2+c[i]-'a'+1)%mod2; 40 int q=read(),l,r,mid; 41 while(q--) 42 { 43 a=read(),b=read(),l=ans=1,r=n-max(a,b)+1; 44 if(c[a]!=c[b]) {puts("0");continue;} 45 while(r>=l) 46 { 47 mid=(l+r)>>1; 48 if(Check(mid)) ans=mid,l=mid+1; 49 else r=mid-1; 50 } 51 printf("%d\n",ans); 52 } 53 }
T2 线段求交
题目大意：
求两个线段的交点
思路：
使用计算几何
1 #include<iostream> 2 #include<cmath> 3 #include<algorithm> 4 #include<cstdio> 5 #include<cstring> 6 #include<cstdlib> 7 #include<queue> 8 #include<vector> 9 #include<set> 10 #define MAXN 11 #define ll long long 12 #define inf 2139062143 13 using namespace std; 14 inline int read() 15 { 16 int x=0,f=1;char ch=getchar(); 17 while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();} 18 while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();} 19 return x*f; 20 } 21 int n,ans; 22 struct Point 23 { 24 double x,y; 25 Point operator - (const Point &a) const 26 { 27 return (Point){x-a.x,y-a.y}; 28 } 29 double operator ^ (const Point &a) const 30 { 31 return x*a.y-y*a.x; 32 } 33 }; 34 struct Segment{Point st,ed;}g,h; 35 double Area(Point a,Point b,Point c){return (b-a)^(c-a);} 36 bool Intersec(Point a,Point b,Point c,Point d) 37 { 38 if(max(a.x,b.x)<min(c.x,d.x))return false; 39 if(max(a.y,b.y)<min(c.y,d.y))return false; 40 if(max(c.x,d.x)<min(a.x,b.x))return false; 41 if(max(c.y,d.y)<min(a.y,b.y))return false; 42 if(Area(c,b,a)*Area(b,d,a)<0)return false; 43 if(Area(a,d,c)*Area(d,b,c)<0)return false; 44 return true; 45 } 46 int main() 47 { 48 freopen("intersect.in","r",stdin); 49 freopen("intersect.out","w",stdout); 50 int T=read(); 51 while(T--) 52 { 53 n=read(),ans=0; 54 g.st.x=read()*1.0,g.st.y=read()*1.0,g.ed.x=read()*1.0,g.ed.y=read()*1.0; 55 for(int i=1;i<=n;i++) 56 { 57 h.st.x=read()*1.0,h.st.y=read()*1.0,h.ed.x=read()*1.0,h.ed.y=read()*1.0; 58 if(Intersec(g.st,g.ed,h.st,h.ed)){ans=1;continue;} 59 } 60 if(ans) puts("YES"); 61 else puts("NO"); 62 } 63 }
T3 spaceship
题目大意：
你在点1处可以带一些油，每条道路有一个额外的权值c，表示当你剩余至少c的油的时候才能走这条路，每条路需要费不同的油
求1到n至少要带多少油
思路：
反跑dij 每次修改dis数组的时候dis[to[i]与max(dis[x]+val[i],limit[i])取min即可
1 #include<iostream> 2 #include<cmath> 3 #include<algorithm> 4 #include<cstdio> 5 #include<cstring> 6 #include<cstdlib> 7 #include<queue> 8 #include<vector> 9 #include<set> 10 #define MAXN 100100 11 #define ll long long 12 #define inf 2139062143 13 using namespace std; 14 inline int read() 15 { 16 int x=0,f=1;char ch=getchar(); 17 while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();} 18 while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();} 19 return x*f; 20 } 21 int n,m,fst[MAXN],nxt[MAXN<<1],to[MAXN<<1],val[MAXN<<1],lmt[MAXN<<1],cnt; 22 void add(int u,int v,int w,int c){nxt[++cnt]=fst[u],fst[u]=cnt,to[cnt]=v,val[cnt]=w,lmt[cnt]=c;} 23 struct node 24 { 25 int x,dis; 26 bool operator < (const node &a) const 27 { 28 return dis>a.dis; 29 } 30 }; 31 int dis[MAXN],vis[MAXN]; 32 priority_queue <node> q; 33 void dij() 34 { 35 memset(dis,127,sizeof(dis)); 36 dis[n]=0;q.push((node){n,0});int x,d; 37 while(!q.empty()) 38 { 39 x=q.top().x,d=q.top().dis;q.pop(); 40 if(vis[x]) continue; 41 vis[x]=1; 42 for(int i=fst[x];i;i=nxt[i]) 43 if(dis[to[i]]>max(d+val[i],lmt[i])){dis[to[i]]=max(d+val[i],lmt[i]);q.push((node){to[i],dis[to[i]]});} 44 } 45 if(dis[1]>=inf) puts("-1"); 46 else printf("%d",dis[1]); 47 } 48 int main() 49 { 50 freopen("spaceship.in","r",stdin); 51 freopen("spaceship.out","w",stdout); 52 n=read(),m=read();int a,b,c,d; 53 while(m--){a=read(),b=read(),c=read(),d=read();add(a,b,c,d);add(b,a,c,d);} 54 dij(); 55 }