T1
[数学]
期望得分：100
先通分，求出分子的最小公倍数，再讨论跟共同的分母B*D的关系即可。
【code】
#include<bits/stdc++.h> using namespace std; #define ll long long #define File "tile" inline void file(){ freopen(File".in","r",stdin); freopen(File".out","w",stdout); } inline int read(){ int x = 0,f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){if(ch=='-')f = -1; ch = getchar();} while(ch >= '0' && ch <= '9'){x = (x<<1) + (x<<3) + ch-'0'; ch = getchar();} return x*f; } //const int mxn = ; int T,A,B,C,D; ll gcd(ll x,ll y){ return y ? gcd(y,x%y) : x; } int main(){ file(); T = read(); while(T--){ A = read(),B = read(),C = read(),D = read(); ll t1 = A*D,t2 = C*B; ll gcd_ = gcd(t1,t2); ll mul_ = t1*t2; ll lcm_ = mul_/gcd_; ll tmp = gcd(lcm_,B*D); if(tmp == B*D) printf("%lld ",lcm_/(B*D)); else printf("%lld/%lld ",lcm_/tmp,(B*D)/tmp); } return 0; }
T2
[？]
枚举所有边，判断是否满足它连接的两个点度数分别为2和3，计数。
还要记录度数为3的点与之相连非当前枚举到的边的最长长度+当前枚举到的边的长度+与度数为2的点相连的最长的边。
为了求最长长度，需要记录每个点的连边中前三大的。
【code】
#include<bits/stdc++.h> using namespace std; #define ll long long #define File "question" inline void file(){ freopen(File".in","r",stdin); freopen(File".out","w",stdout); } inline int read(){ int x = 0,f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){if(ch=='-')f = -1; ch = getchar();} while(ch >= '0' && ch <= '9'){x = (x<<1) + (x<<3) + ch-'0'; ch = getchar();} return x*f; } const int mxn = 2e5 + 5; int n; struct edge{ int x,y,v; }e[mxn<<1]; int deg[mxn]; int mx1[mxn],mx2[mxn],mx3[mxn]; int to1[mxn],to2[mxn]; inline void prewor(int x,int y,int z){ if(z >= mx1[x]){ mx3[x] = mx2[x],mx2[x] = mx1[x],mx1[x] = z; to2[x] = to1[x],to1[x] = y; }else if(z >= mx2[x]){ mx3[x] = mx2[x],mx2[x] = z; to2[x] = y; }else if(z > mx3[x]) mx3[x] = z; } ll cnt,mx; inline void wor(int x,int y,int z){ if(deg[x] >= 3 && deg[y] >= 2) cnt += (deg[x]-1)*(deg[x]-2)*(deg[y]-1)/2; if(y==to1[x] && mx3[x]){ if(x==to1[y]) if(mx2[y]) mx = max(mx,mx2[y] + z + mx2[x] + mx3[x]); //该边为最长边 else if(mx1[y]) mx = max(mx,mx2[x] + mx3[x] + mx1[y] + z); }else if(y==to2[x] && mx3[x]){ if(x==to1[y]) if(mx2[y]) mx = max(mx,mx1[x] + z + mx3[x] + mx2[y]); else if(mx1[y]) mx = max(mx,mx1[y] + z + mx1[x] + mx3[x]); }else { } } int main(){ // file(); n = read(); for(int i = 1;i < n; ++i){ int x = read(),y = read(),v = read(); e[i] = (edge){x,y,v}; deg[x]++,deg[y]++; prewor(x,y,v),prewor(y,x,v); } for(int i = 1;i < n; ++i){ } printf("%lld %lld ",cnt,mx); return 0; } /* 7 1 3 2 2 3 1 3 5 1 5 4 2 4 6 3 5 7 3 */
T3
[搜索+剪枝]
每个水管可以看做是2+2（分别朝不同的方向）或者3+1。
直接搜索+剪枝。
根据当前点到终点的曼哈顿距离，可以求出这个点到终点的最少步数。
如果当前点加上到达终点的最小步数，还是没有ans优，直接剪枝。
注意代码细节处理。
【code】
#include<bits/stdc++.h> using namespace std; #define ll long long #define File "plumber" inline void file(){ freopen(File".in","r",stdin); freopen(File".out","w",stdout); } inline int read(){ int x = 0,f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){if(ch=='-')f = -1; ch = getchar();} while(ch >= '0' && ch <= '9'){x = (x<<1) + (x<<3) + ch-'0'; ch = getchar();} return x*f; } const int mxn = 25; int X,Y,Z,m; int st_x,st_y,st_z,st_k; int ed_x,ed_y,ed_z,ed_k; char st_tp,ed_tp; bool v[mxn][mxn][mxn],vis[mxn][mxn][mxn]; const int dx[6] = {1,-1,0,0,0,0}; const int dy[6] = {0,0,1,-1,0,0}; const int dz[6] = {0,0,0,0,1,-1}; inline bool ok(int x,int y,int z){ return x>0&&x<=X&&y>0&&y<=Y&&z>0&&z<=Z&&!v[x][y][z]&&!vis[x][y][z]; } int ans = 13; void dfs(int x,int y,int z,int d,int cnt){ int dis = (abs(ed_x-x)+abs(ed_y-y)+abs(ed_z-z))/4; if(dis + cnt >= ans) return; if(x==ed_x && y==ed_y && z==ed_z){ if(d==ed_k) ans = cnt; return; } if(ok(x+dx[d],y+dy[d],z+dz[d]) && ok(x+dx[d]*2,y+dy[d]*2,z+dz[d]*2)){ vis[x+dx[d]][y+dy[d]][z+dz[d]] = vis[x+dx[d]*2][y+dy[d]*2][z+dz[d]*2] = 1; int x_ = x+dx[d]*2,y_ = y+dy[d]*2,z_ = z+dz[d]*2; for(int i = 0;i < 6; ++i){//2+2 if((i/2) != (d/2)){ if(ok(x_+dx[i],y_+dy[i],z_+dz[i]) && ok(x_+dx[i]*2,y_+dy[i]*2,z_+dz[i]*2)){ vis[x_+dx[i]][y_+dy[i]][z_+dz[i]] = vis[x_+dx[i]*2][y_+dy[i]*2][z_+dz[i]*2] = 1; dfs(x_+dx[i]*2,y_+dy[i]*2,z_+dz[i]*2,i,cnt+1); vis[x_+dx[i]][y_+dy[i]][z_+dz[i]] = vis[x_+dx[i]*2][y_+dy[i]*2][z_+dz[i]*2] = 0; } } } if(ok(x+dx[d]*3,y+dy[d]*3,z+dz[d]*3)){ vis[x+dx[d]*3][y+dy[d]*3][z+dz[d]*3] = 1; int _x = x+dx[d]*3,_y = y+dy[d]*3,_z = z+dz[d]*3; for(int i = 0;i < 6; ++i){ if((i/2) != (d/2)){ if(ok(_x+dx[i],_y+dy[i],_z+dz[i])){ vis[_x+dx[i]][_y+dy[i]][_z+dz[i]] = 1; dfs(_x+dx[i],_y+dy[i],_z+dz[i],i,cnt+1); vis[_x+dx[i]][_y+dy[i]][_z+dz[i]] = 0; } } } vis[x+dx[d]*3][y+dy[d]*3][z+dz[d]*3] = 0; } vis[x+dx[d]][y+dy[d]][z+dz[d]] = vis[x+dx[d]*2][y+dy[d]*2][z+dz[d]*2] = 0; } } int main(){ file(); X = read(),Y = read(),Z = read(),m = read(); st_x = read(),st_y = read(),st_z = read(),st_tp = getchar(); if(st_tp=='x') st_k = (st_x==1)?0:1; if(st_tp=='y') st_k = (st_y==1)?2:3; if(st_tp=='z') st_k = (st_z==1)?4:5; ed_x = read(),ed_y = read(),ed_z = read(),ed_tp = getchar(); if(ed_tp=='x') ed_k = (ed_x==1)?1:0; if(ed_tp=='y') ed_k = (ed_y==1)?3:2; if(ed_tp=='z') ed_k = (ed_z==1)?5:4; for(int i = 1;i <= m; ++i){ int x = read(),y = read(),z = read(); v[x][y][z] = 1; } dfs(st_x-dx[st_k],st_y-dy[st_k],st_z-dz[st_k],st_k,0); // printf("%d ",ans); ans < 13 ? printf("%d ",ans) : puts("impossible"); return 0; } /* 5 4 3 1 3 1 1 z 1 4 3 x 2 3 3 */