2015 CCPC 个人题解

C-The Battle of Chibi

每次问长度为k的上升子序列有多少个，可以直接树状数组然后DP。

#include
//#define int long long
#define lb(x) (x&-x)
using namespace std;
typedef long long ll;
const int N=2007;
const ll mod=1e9+7;inline int read(){int x=0;char ch=getchar();while(!isdigit(ch)) ch=getchar();while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();return x;
}int rx[N*N],ry[N*N],rz[N*N],top;vectorvt;
int n,m,a[N],rk[N];
ll dp[N][N],tr[N][N<<1];
inline void update(int id,int x,int w){if(!x) return;for(;xll res=0;for(;x;x-=lb(x)) res=(res+tr[id][x])%mod;return res;
}void solve(){vt.clear();top=0;n=read();m=read();vt.emplace_back(0);for(int i=1;i<=n;++i) a[i]=read(),vt.emplace_back(a[i]);sort(vt.begin(),vt.end());for(int i=1;i<=n;++i) rk[i]=lower_bound(vt.begin(),vt.end(),a[i])-vt.begin()+1;rk[0]=1;update(0,1,1);for(int i=1;i<=n;++i){for(int k=0;kdp[i][k+1]=(dp[i][k+1]+query(k,rk[i]-1))%mod;update(k+1,rk[i],dp[i][k+1]);++top;rx[top]=k+1;ry[top]=rk[i];rz[top]=dp[i][k+1];}}ll ans=0;for(int i=1;i<=n;++i) ans+=dp[i][m],ans%=mod;for(int i=0;i<=n;++i) for(int j=0;j<=m;++j) dp[i][j]=0;while(top){update(rx[top],ry[top],-rz[top]);--top;}update(0,1,-1);printf("%lld\n",ans);
}signed main(){int T=1;T=read();for(int i=1;i<=T;++i){printf("Case #%d: ",i);solve();}
}
/*
3
3 2
1 2 3
3 2
3 2 1
5 2
1 2 1 2 33
8 2
42 42 50 10 89 98 12 4
4 2
77 55 95 35
3 2
15 27 85*/

D-Pick The Sticks

可以转化为背包问题，dp[i][j][0]dp[i][j][0]dp[i][j][0]表示不放在边上，使用长度jjj的位置能在前iii个中最高取得多少价值，同理，dp[i][j][1]和dp[i][j][2]dp[i][j][1]和dp[i][j][2]dp[i][j][1]和dp[i][j][2]分别表示把当前物品中心放在最左边和最右边的情况，第一维滚动掉，就可以了。然后要特判一根肯定是可以放的。

#include
#define int long long 
using namespace std;
const int N=4007;
int n,L,w[N],v[N],dp[N][5];void solve(int t){cin>>n>>L;L<<=1;int ans=0;for(int i=1;i<=n;++i){cin>>w[i]>>v[i];ans=max(ans,v[i]); //至少能放一根上去 w[i]<<=1;}memset(dp,0,sizeof(dp));for(int i=1;i<=n;++i){for(int j=L;j>=0;--j){if(j>=w[i]){for(int k=0;k<=2;++k) dp[j][k]=max(dp[j][k],dp[j-w[i]][k]+v[i]);}if(j>=w[i]/2){dp[j][2]=max(dp[j][2],dp[j-w[i]/2][1]+v[i]);dp[j][1]=max(dp[j][1],dp[j-w[i]/2][0]+v[i]);}}}for(int j=L;j>=0;--j){ans=max(ans,dp[j][0]);ans=max(ans,dp[j][1]);ans=max(ans,dp[j][2]);}cout<<"Case #"<ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int T=1;cin>>T;for(int i=1;i<=T;++i){solve(i);}return 0;
}

E-Ba Gua Zhen

找最大异或和回路，应该也是板子，使用高斯消元或者线性基都能做。

#include
#define int long long
#define pii pair
#define mk make_pair
#define F first
#define S second 
using namespace std;
const int N=5e5+7;vectorG[N];
vectorans;
int n,m,Xor[N],vis[N];void dfs(int x,int f,int val){if(vis[x]){ans.emplace_back(val^Xor[x]);return;}vis[x]=1;for(auto p:G[x]){int to=p.F,w=p.S;if(to==f) continue;if(!vis[to]) Xor[to]=val^w;dfs(to,x,val^w);}
}void solve(int t){cin>>n>>m;ans.clear();memset(vis,0,sizeof(vis));memset(Xor,0,sizeof(Xor));for(int i=1;i<=n;++i) G[i].clear();for(int i=1,u,v,w;i<=m;++i){cin>>u>>v>>w;G[u].emplace_back(mk(v,w));G[v].emplace_back(mk(u,w));}dfs(1,-1,0);int k=0,j;for(int i=60;i>=0;--i){for(j=k;jios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int T=1;cin>>T;for(int t=1;t<=T;++t){solve(t);}return 0;
} 

G-Ancient Go

直接枚举每一个o的位置，找其连通块，并记录这个连通块的气（就是围棋里还没堵住的位置），如果是1的话就能吃掉了。模拟即可。

#include
//#define int long long
using namespace std;char mp[20][20];
int vis[20][20],ct[20][20],idx,res=0;inline void check(int x,int y,int id){if(vis[x][y]||x<1||x>9||y<1||y>9) return;vis[x][y]=id;if(mp[x][y-1]=='o') check(x,y-1,id);if(mp[x][y+1]=='o') check(x,y+1,id);if(mp[x+1][y]=='o') check(x+1,y,id);if(mp[x-1][y]=='o') check(x-1,y,id);if(mp[x][y-1]=='.'&&!ct[x][y-1]) ct[x][y-1]=1,++res;if(mp[x][y+1]=='.'&&!ct[x][y+1]) ct[x][y+1]=1,++res;if(mp[x-1][y]=='.'&&!ct[x-1][y]) ct[x-1][y]=1,++res;if(mp[x+1][y]=='.'&&!ct[x+1][y]) ct[x+1][y]=1,++res;
}void solve(){idx=0;memset(vis,0,sizeof(vis));for(int i=1;i<=9;++i){for(int j=1;j<=9;++j){cin>>mp[i][j];}}for(int i=0;i<=10;++i) mp[i][0]=mp[0][i]=mp[10][i]=mp[i][10]='x';for(int i=1;i<=9;++i){for(int j=1;j<=9;++j){res=0;if(!vis[i][j]&&mp[i][j]=='o'){memset(ct,0,sizeof(ct));check(i,j,++idx);if(res==1){cout<<"Can kill in one move!!!\n";return;}} }}cout<<"Can not kill in one move!!!\n";
}signed main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int T=1;cin>>T;for(int i=1;i<=T;++i){cout<<"Case #"<

#include
#define int long long
#define pii pair
#define mk make_pair
using namespace std;char mp[5][5],tmp[5][5];
int visx[5][5],visy[5][5];
int num;
vectorvt;bool check(){for(int i=1;i<=4;++i){  //检查小格 for(int j=1;j<=4;j+=2){for(int k=1;k<=4;k+=2){if(tmp[j][k]==tmp[j][k+1]) return false;if(tmp[j][k]==tmp[j+1][k]) return false;if(tmp[j][k]==tmp[j+1][k+1]) return false;if(tmp[j+1][k]==tmp[j][k+1]) return false;if(tmp[j+1][k]==tmp[j+1][k+1]) return false;if(tmp[j][k+1]==tmp[j+1][k+1]) return false;}}}return true;
}void dfs(int stp){if(stp==num){  //找到答案 if(!check()) return;for(int i=1;i<=4;++i){for(int j=1;j<=4;++j){cout<if(!visx[x][i]&&!visy[y][i]){visx[x][i]=1;visy[y][i]=1;tmp[x][y]=i+'0';dfs(stp+1);visx[x][i]=0;visy[y][i]=0;}}
}void solve(){vt.clear();num=0;memset(visx,0,sizeof(visx));memset(visy,0,sizeof(visy));for(int i=1;i<=4;++i){for(int j=1;j<=4;++j){cin>>mp[i][j];tmp[i][j]=mp[i][j];}}for(int i=1;i<=4;++i){for(int j=1;j<=4;++j){if(mp[i][j]=='*'){++num;vt.emplace_back(mk(i,j));}else{visx[i][mp[i][j]-'0']=1;visy[j][mp[i][j]-'0']=1;}}}dfs(0);
}signed main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int T=1;cin>>T;for(int i=1;i<=T;++i){cout<<"Case #"<

#include
#define int long long
using namespace std;
const int N=4007;int pre[2][N],ppre[2][N],suf[2][N],ssuf[2][N];
int n,a[2][N],dp[2][N]; int calc(int l,int r,int f){ //cost函数 f^=1;int mid=((l+r)>>1);if(l==1&&r==n) return 0x3f3f3f3f3f3f3f3f;if(l==1) return ppre[f][r];if(r==n) return ssuf[f][l];int res=ssuf[f][l]-ssuf[f][mid+1]-(mid-l+1)*suf[f][mid+1];res+=ppre[f][r]-ppre[f][mid]-(r-mid)*pre[f][mid]; return res;
}void init(){memset(a,0,sizeof(a));memset(dp,0x3f,sizeof(dp));memset(pre,0,sizeof(pre));memset(suf,0,sizeof(suf));memset(ppre,0,sizeof(ppre));memset(ssuf,0,sizeof(ssuf));
}void solve(int t){cin>>n;init();for(int i=1;i<=n;++i) cin>>a[0][i]>>a[1][i];dp[0][0]=dp[1][0]=0;for(int i=1;i<=n;++i){for(int k=0;k<=1;++k){pre[k][i]=pre[k][i-1]+a[k][i];ppre[k][i]=ppre[k][i-1]+pre[k][i];}}for(int i=n;i;--i){for(int k=0;k<=1;++k){suf[k][i]=suf[k][i+1]+a[k][i];ssuf[k][i]=ssuf[k][i+1]+suf[k][i];}}for(int i=1;i<=n;++i){for(int k=0;k<=1;++k){for(int j=0;jdp[k][i]=min(dp[k][i],dp[k^1][j]+calc(j+1,i,k));}}}cout<<"Case #"<ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int T=1;cin>>T;for(int i=1;i<=T;++i){solve(i);}return 0;
}

#include
#define int long long
using namespace std;void solve(){int n;cin>>n;cout<<2*n-1<<"\n";
}signed main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int T=1;cin>>T;for(int i=1;i<=T;++i){cout<<"Case #"<