solve 5/11
Code:zz
Thinking :zz
题解待补
#include<iostream> #include<stdio.h> #include<stdlib.h> #include<string.h> #include<string> #include<math.h> #include<cmath> #include<time.h> #include<map> #include<set> #include<vector> #include<queue> #include<algorithm> #include<numeric> #include<stack> #include<bitset> #include<unordered_map> const int maxn = 0x3f3f3f3f; const double EI = 2.71828182845904523536028747135266249775724709369995957496696762772407663035354594571382178525166427; const double PI = 3.141592653589793238462643383279; using namespace std; struct s { int a,b; }z[2020],ans[2020]; vector<int>ve[2020]; int c[2020],hi[2020],d[2020]; int main(void) { int n,m,i,si,j,pos,cnt; while(~scanf("%d %d",&n,&m)) { memset(c,0,sizeof(c)); memset(d,0,sizeof(d)); memset(hi,0,sizeof(hi)); for(i = 0;i <= n;i++) { ve[i].clear(); } for(i = 0;i < m;i++) { scanf("%d %d",&z[i].a,&z[i].b); ve[z[i].a].push_back(z[i].b); ve[z[i].b].push_back(z[i].a); } /*for(i = 1;i <= n;i++) { printf("%d: ",i); for(j = 0;j < ve[i].size();j++) { printf("%d ",ve[i][j]); } printf("\n"); }*/ queue<int>q; q.push(1); cnt = 0; c[1] = 1; d[1] = 1; hi[1] = 1; while(!q.empty()) { pos = q.front(); q.pop(); //c[pos] = cnt++; //printf("%d %d\n",c[pos],d[pos]); ans[pos].a = c[pos]; ans[pos].b = d[pos]; si = ve[pos].size(); //printf("pos = %d : ",pos); for(j = 0;j < si;j++) { //printf(" %d ",ve[pos][j]); if(!c[ve[pos][j]]) { c[ve[pos][j]] = c[pos] + 1; d[ve[pos][j]] = hi[c[ve[pos][j]]] + 1; hi[c[ve[pos][j]]]++; q.push(ve[pos][j]); } } //printf("\n"); } for(i = 1;i <= n;i++) { printf("%d %d\n",ans[i].a,ans[i].b); } } return 0; }
Code:zz
Thinking:zz
题解待补
#include<iostream> #include<stdio.h> #include<stdlib.h> #include<string.h> #include<string> #include<math.h> #include<cmath> #include<time.h> #include<map> #include<set> #include<vector> #include<queue> #include<algorithm> #include<numeric> #include<stack> #include<bitset> #include<unordered_map> const int maxn = 0x3f3f3f3f; const double EI = 2.71828182845904523536028747135266249775724709369995957496696762772407663035354594571382178525166427; const double PI = 3.141592653589793238462643383279; using namespace std; long long c[100010]; int main(void) { int n, m, i; long long ans, pos; while (~scanf("%d %d", &n, &m)) { priority_queue<long long>q; for (i = 0; i < n; i++) { scanf("%lld", c + i); q.push(c[i]); } while (m > 0) { pos = q.top(); //printf("%d\n",pos); q.pop(); if (pos == 1 || m == 0) { break; } pos /= 2; q.push(pos); m--; } ans = 0; while (!q.empty()) { pos = q.top(); q.pop(); ans += pos; } printf("%lld\n", ans); } return 0; }
搜索，待补，比赛的时候kk和pai爷提出了两个能ac的想法，但踏马的就是没写。
Code：kk
Thinking：pai爷 kk
状压dp
$f[ i ] [ s ] [ j ] +=f[ i-1 ] [ s' ][ k ]$
表示第 i 个位置，状态为s（二进制），第i个位置填j的状态，然后就从题目给出的限制条件进行转移，看代码即可理解。
#include<bits/stdc++.h> #define CLR(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; const ll p=1e9+7; ll f[18][(1<<15)+10][18]; int n; char op[20]; int main(){ while(cin>>n){ scanf("%s",op+1); CLR(f,0); for(int i=1;i<=n;i++) { f[1][1<<(i-1)][i]=1; } int tmp=0,tot=0; for(int i=2;i<=n;i++) { for(int j=1;j<=n;j++) { for(int s=0;s<(1<<n);s++) { tmp=s;//判断数字个数 合法 tot=0; while(tmp>0){ tot+=tmp%2; tmp>>=1; } if(tot!=i)continue; if(( s&(1<<(j-1)) )==0)continue;//没有了 j for(int k=1;k<=n;k++) { if(k==j)continue; if((1<<(k-1))&s==0)continue;//没有k if(op[i-1]=='0'){ if((j/k!=2||j%k!=0) && (k/j!=2||k%j!=0) ){ f[i][s][j]+=f[i-1][s^(1<<(j-1))][k]%p; f[i][s][j]%=p; } }else{ if((j/k==2&&j%k==0 )|| (k/j==2&&k%j==0)){ f[i][s][j]+=f[i-1][s^(1<<(j-1))][k]%p; f[i][s][j]%=p; } } } } } } ll ans=0; // printf("debug:%d\n",(1<<n)-1); for(int j=1;j<=n;j++) { ans=(ans+f[n][(1<<n)-1][j]%p)%p; } printf("%lld\n",ans); } }
Code: kk
Thinking :pai爷 kk
添加边，其实就是连成了一棵树，就是！一！棵！树！没有其他的有的没的，树形dfs一下，记录一下每个节点的儿子树，父边被走过的总数量就是 $son[ i ]* (n-son [i ])$
#include<bits/stdc++.h> #define CLR(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; const int maxn=1e6+10; int head[maxn],tot; struct edge{ int to,Next; }a[maxn<<2]; ll son[maxn]; int n,m,x,y,u,v; ll zo; ll ans=0; const ll p=1e9+7; void init(){ CLR(son,0); CLR(head,-1); tot=0; ans=0; } void addv(int u,int v){ a[++tot].to=v; a[tot].Next=head[u]; head[u]=tot; } void dfs(int u,int fa){ son[u]=1; for(int i=head[u];i!=-1;i=a[i].Next) { int v=a[i].to; if(v==fa)continue; dfs(v,u); son[u]+=son[v]; } if(fa!=-1){ ans=(ans+son[u]*(zo-son[u])%p)%p; } } int main(){ while(cin>>n>>m) { init(); zo=((ll)n*m); for(int i=1;i<n;i++) { scanf("%d%d",&u,&v); addv(u,v); addv(v,u); for(int j=2;j<=m;j++) { addv((j-1)*n+u,(j-1)*n+v); addv((j-1)*n+v,(j-1)*n+u); } } m--; while(m--) { scanf("%d%d%d%d",&x,&y,&u,&v); addv((x-1)*n+u,(y-1)*n+v); addv((y-1)*n+v,(x-1)*n+u); } dfs(1,-1); printf("%lld\n",ans); } }
线段树，待补，好巧妙的思想。
Code：zz
Thinking：zz
题解待补
#include<bits/stdc++.h> #define CLR(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; const int maxn = 0x3f3f3f3f; using namespace std; struct Point { long long x; long long y; }; typedef struct Point point; long long multi(point p0, point p1, point p2) { //return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y); if((p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y) > 0) { return 1; } if((p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y) < 0) { return -1; } return 0; } bool isIntersected(point s1, point e1, point s2, point e2) { return (max(s1.x, e1.x) >= min(s2.x, e2.x)) && (max(s2.x, e2.x) >= min(s1.x, e1.x)) && (max(s1.y, e1.y) >= min(s2.y, e2.y)) && (max(s2.y, e2.y) >= min(s1.y, e1.y)) && (multi(s1, s2, e1)*multi(s1, e1, e2)>=0) && (multi(s2, s1, e2)*multi(s2, e2, e1)>=0); } struct s { int a,b; }z[2020]; int main(void) { int n,m,i,j,ans; while(~scanf("%d %d",&n,&m)) { point po[2020]; for(i = 0;i < m;i++) { scanf("%d %d",&z[i].a,&z[i].b); } for(i = 1;i <= n;i++) { scanf("%lld %lld",&po[i].x,&po[i].y); } ans = 0; for(i = 0;i < m;i++) { for(j = i + 1;j < m;j++) { if(po[z[i].a].x == po[z[j].a].x && po[z[i].a].y == po[z[j].a].y || po[z[i].a].x == po[z[j].b].x && po[z[i].a].y == po[z[j].b].y || po[z[i].b].x == po[z[j].a].x && po[z[i].b].y == po[z[j].a].y || po[z[i].b].x == po[z[j].b].x && po[z[i].b].y == po[z[j].b].y) { if((po[z[i].a].x - po[z[i].b].x) * (po[z[j].a].y - po[z[j].b].y) == (po[z[i].a].y - po[z[i].b].y) * (po[z[j].a].x - po[z[j].b].x)) { if((po[z[i].a].x == po[z[j].a].x && po[z[i].a].y == po[z[j].a].y && (po[z[i].b].x - po[z[i].a].x) * (po[z[j].b].x - po[z[i].a].x) >= 0 || po[z[i].a].x == po[z[j].b].x && po[z[i].a].y == po[z[j].b].y && (po[z[i].b].x - po[z[i].a].x) * (po[z[j].a].x - po[z[i].a].x) >= 0 || po[z[i].b].x == po[z[j].a].x && po[z[i].b].y == po[z[j].a].y && (po[z[i].a].x - po[z[i].b].x) * (po[z[j].b].x - po[z[i].b].x) >= 0 ||po[z[i].b].x == po[z[j].b].x && po[z[i].b].y == po[z[j].b].y && (po[z[i].a].x - po[z[i].b].x) * (po[z[j].a].x - po[z[i].b].x) >= 0) && (po[z[i].a].y == po[z[i].b].y)) { ans++; } else if((po[z[i].a].x == po[z[j].a].x && po[z[i].a].y == po[z[j].a].y && (po[z[i].b].y - po[z[i].a].y) * (po[z[j].b].y - po[z[i].a].y) >= 0 || po[z[i].a].x == po[z[j].b].x && po[z[i].a].y == po[z[j].b].y && (po[z[i].b].y - po[z[i].a].y) * (po[z[j].a].y - po[z[i].a].y) >= 0 || po[z[i].b].x == po[z[j].a].x && po[z[i].b].y == po[z[j].a].y && (po[z[i].a].y - po[z[i].b].y) * (po[z[j].b].y - po[z[i].b].y) >= 0 ||po[z[i].b].x == po[z[j].b].x && po[z[i].b].y == po[z[j].b].y && (po[z[i].a].y - po[z[i].b].y) * (po[z[j].a].y - po[z[i].b].y) >= 0) && (po[z[i].a].x == po[z[i].b].x)) { ans++; } else if((po[z[i].a].x == po[z[j].a].x && po[z[i].a].y == po[z[j].a].y && (po[z[i].b].y - po[z[i].a].y) * (po[z[j].b].y - po[z[i].a].y) >= 0 || po[z[i].a].x == po[z[j].b].x && po[z[i].a].y == po[z[j].b].y && (po[z[i].b].y - po[z[i].a].y) * (po[z[j].a].y - po[z[i].a].y) >= 0 || po[z[i].b].x == po[z[j].a].x && po[z[i].b].y == po[z[j].a].y && (po[z[i].a].y - po[z[i].b].y) * (po[z[j].b].y - po[z[i].b].y) >= 0 ||po[z[i].b].x == po[z[j].b].x && po[z[i].b].y == po[z[j].b].y && (po[z[i].a].y - po[z[i].b].y) * (po[z[j].a].y - po[z[i].b].y) >= 0)) { ans++; } } continue; } bool flag = isIntersected(po[z[i].a],po[z[i].b],po[z[j].a],po[z[j].b]); if(flag) { //printf("%d %d %d %d\n",z[i].a,z[i].b,z[j].a,z[j].b); ans++; } } } printf("%d\n",ans); } return 0; }
赛后总结：
kk：今天开局以为几何题是板子题，眉头一皱发现事情并不简单，就看h题去了，结果智障的没有思路，pai爷一点拨就想通了，后来f题也在和pai爷的合作下写完了，感觉今天状态有点迷吧，还好代码都是一发ac的，以后不要再在h题这种简单题上卡了，加油。
pai爷：今天浑水摸鱼，想尝试几道数列的题，奈何水平不够都咕了。
zz：