QAQ(蒟蒻一枚),居然全打暴力。
T1能看出来是DP,但实在想不到那个奇技淫巧,果断暴搜(20)。
T2应该比较简单,写完正解跟暴力对拍,错了QAQ,直接交了暴力。
T3压根没往floyd想,交暴力。
T1(cogs 1292.打砖块)
暴力:
![](http://img.e-com-net.com/image/info8/b8d97b5613f94ed2ba791cad57d0b2ed.gif)
![](http://img.e-com-net.com/image/info8/2f88dd3f1cd145f59c0e47b51acdbd4b.gif)
1 #include2 #include 3 #define mysister 4 using namespace std; 5 const int maxn=50; 6 int read(){ 7 int x=0,f=1;char ch=getchar(); 8 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 9 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 10 return x*f; 11 } 12 void Emine(){ 13 freopen("brike.in","r",stdin); 14 freopen("brike.out","w",stdout); 15 } 16 int n,m,a[maxn][maxn],ans=0,vis[maxn][maxn]; 17 void dfs(int cc,int aa,int bb){ 18 if(cc==m){ 19 int anss=0; 20 for(int i=0;i i) 21 for(int j=0;j j) 22 if(vis[i][j]) 23 anss+=a[i][j]; 24 ans=max(anss,ans); 25 return; 26 } 27 if(bb>=n-aa||aa>=n) return; 28 for(int i=bb;i i) 29 if(aa==0||(vis[aa-1][i]&&vis[aa-1][i+1])){ 30 vis[aa][i]=1; 31 dfs(cc+1,aa,i+1);dfs(cc+1,aa+1,0); 32 vis[aa][i]=0; 33 } 34 } 35 int main(){ 36 Emine(); 37 n=read(),m=read(); 38 for(int i=0;i i) 39 for(int j=0;j j) 40 a[i][j]=read(); 41 dfs(0,0,0); 42 printf("%d\n",ans); 43 return 0; 44 }
正解:
![](http://img.e-com-net.com/image/info8/b8d97b5613f94ed2ba791cad57d0b2ed.gif)
![](http://img.e-com-net.com/image/info8/2f88dd3f1cd145f59c0e47b51acdbd4b.gif)
1 //f[i][j][k]表示截止到第i行,总共已经选j个砖块,其中第i行已经选了前k个砖块的最大值。 2 #include3 #include 4 #include 5 #include 6 #define maxn 55 7 using namespace std; 8 void Emine(){ 9 freopen("brike.in","r",stdin); 10 freopen("brike.out","w",stdout); 11 } 12 int n,m,v[maxn][maxn],s[maxn],sum[maxn][maxn],ans,f[maxn][10*maxn][maxn]; 13 int main(){ 14 Emine(); 15 scanf("%d%d",&n,&m); 16 for(int i=1;i<=n;i++)for(int j=i;j<=n;j++) scanf("%d",&v[j][i]); 17 for(int i=1;i<=n;i++)for(int j=1;j<=i;j++) sum[i][j]=sum[i][j-1]+v[i][j]; 18 for(int i=1;i<=n;i++) s[i]=s[i-1]+i; 19 memset(f,-333,sizeof(f)); f[0][0][0]=0; 20 for(int i=1;i<=n;i++){//截止到第i行 21 for(int j=0;j<=min(m,s[i]);j++){//总共已经选j个砖块 22 for(int k=0;k<=min(i,j);k++){//第i行已经选了前k个砖块 23 for(int p=max(k-1,0);p) 24 f[i][j][k]=max(f[i][j][k],f[i-1][j-k][p]+sum[i][k]); 25 ans=max(ans,f[i][j][k]); 26 } 27 } 28 } 29 printf("%d\n",ans); 30 return 0; 31 }
T2(luogu 1631.序列合并)
暴力:
![](http://img.e-com-net.com/image/info8/b8d97b5613f94ed2ba791cad57d0b2ed.gif)
![](http://img.e-com-net.com/image/info8/2f88dd3f1cd145f59c0e47b51acdbd4b.gif)
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 #define ll long long 8 using namespace std; 9 priority_queue ,greater > q; 10 int n;ll a[100005],b[100005]; 11 int main(){ 12 scanf("%d",&n); 13 for(int i=1;i<=n;i++)scanf("%lld",&a[i]); 14 for(int i=1;i<=n;i++)scanf("%lld",&b[i]); 15 for(int i=1;i<=n;i++) 16 for(int j=1;j<=n;j++) 17 q.push(a[i]+b[j]); 18 for(int i=1;i<=n;i++){ 19 printf("%lld ",q.top()); 20 q.pop(); 21 } 22 return 0; 23 }
正解:
![](http://img.e-com-net.com/image/info8/b8d97b5613f94ed2ba791cad57d0b2ed.gif)
![](http://img.e-com-net.com/image/info8/2f88dd3f1cd145f59c0e47b51acdbd4b.gif)
1 #include2 #include 3 #include 4 #include 5 #define maxn 100005 6 using namespace std; 7 int a[maxn],b[maxn],ans[maxn]; 8 priority_queue <int> q; 9 int main(){ 10 int n;cin>>n; 11 for(int i=1;i<=n;i++)scanf("%d",&a[i]); 12 for(int i=1;i<=n;i++){ 13 scanf("%d",&b[i]); 14 q.push(b[i]+a[1]); 15 } 16 for(int i=1;i<=n;i++) 17 for(int j=2;j<=n;j++){ 18 int temp=b[i]+a[j]; 19 if(temp>q.top()) break; 20 else{ 21 q.pop(); 22 q.push(temp); 23 } 24 } 25 for(int i=1;i<=n;i++){ 26 ans[n-i+1]=q.top(); 27 q.pop(); 28 } 29 for(int i=1;i<=n;i++) 30 cout< " "; 31 return 0; 32 }
T3(cogs 501.最小密度路径)
暴力:
![](http://img.e-com-net.com/image/info8/b8d97b5613f94ed2ba791cad57d0b2ed.gif)
![](http://img.e-com-net.com/image/info8/2f88dd3f1cd145f59c0e47b51acdbd4b.gif)
1 #include2 #include 3 #include 4 #include 5 using namespace std; 6 int cnt,n,m,x,y,q,next[100001],last[100001],to[100001],cost[100001],b[101][101]; 7 double ans,a[101][101]; 8 int read(){ 9 int x=0,f=1;char ch=getchar(); 10 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 11 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 12 return x*f; 13 } 14 void Emine(){ 15 freopen("path.in","r",stdin); 16 freopen("path.out","w",stdout); 17 } 18 void dfs(int num,int sum,int len){ 19 if(num==y){ 20 if((double)sum/(double)len double)sum/(double)len; 21 return ; 22 } 23 int t=last[num]; 24 while(t){ 25 dfs(to[t],sum+cost[t],len+1); 26 t=next[t]; 27 } 28 } 29 int main(){ 30 Emine(); 31 n=read(),m=read(); 32 for(int i=1;i<=m;i++){ 33 x=read(),y=read(); 34 next[++cnt]=last[x],last[x]=cnt,to[cnt]=y,cost[cnt]=read(); 35 } 36 scanf("%d",&q); 37 while(q--){ 38 x=read(),y=read(),ans=100002; 39 if(!b[x][y]){ 40 dfs(x,0,0); 41 a[x][y]=ans,b[x][y]=1; 42 } 43 else ans=a[x][y]; 44 if(ans>=100002) printf("OMG!\n"); 45 else printf("%.3lf\n",ans); 46 } 47 return 0; 48 }
正解:
![](http://img.e-com-net.com/image/info8/b8d97b5613f94ed2ba791cad57d0b2ed.gif)
![](http://img.e-com-net.com/image/info8/2f88dd3f1cd145f59c0e47b51acdbd4b.gif)
1 #include2 #include 3 #include 4 #include 5 #define maxn 55 6 #define maxm 1005 7 #define inf 0x3f3f3f3f 8 using namespace std; 9 int d[maxn][maxn][maxn],n,m,q; 10 double ans[maxn][maxn]; 11 void Emine(){ 12 freopen("path.in","r",stdin); 13 freopen("path.out","w",stdout); 14 } 15 int main(){ 16 Emine(); 17 scanf("%d%d",&n,&m); 18 memset(d,inf,sizeof(d)); 19 for(int i=1;i<=m;i++){ 20 int u,v,w;scanf("%d%d%d",&u,&v,&w); 21 d[u][v][1]=min(d[u][v][1],w); 22 } 23 for(int t=2;t<=n;t++) 24 for(int k=1;k<=n;k++) 25 for(int i=1;i<=n;i++) 26 for(int j=1;j<=n;j++){ 27 if(d[i][k][t-1]==inf||d[k][j][1]==inf) continue; 28 d[i][j][t]=min(d[i][j][t],d[i][k][t-1]+d[k][j][1]); 29 } 30 for(int i=1;i<=n;i++) 31 for(int j=1;j<=n;j++){ 32 ans[i][j]=inf; 33 if(i==j){ans[i][j]=0;continue;} 34 for(int k=1;k<=n;k++) ans[i][j]=min(ans[i][j],(double)d[i][j][k]/(double)k); 35 } 36 scanf("%d",&q); 37 while(q--){ 38 int u,v;scanf("%d%d",&u,&v); 39 if(ans[u][v]>=1e6) printf("OMG!\n"); //被坑QAQ,最大密度为最大边权w/1==1e6; 40 else printf("%.3lf\n",ans[u][v]); 41 } 42 return 0; 43 }