暑训day1解题报告

A - Painting the sticks

因为不能覆盖涂/涂两次，所以就数数有几个三个一块儿就行了。

#include<cstdio>
int a[100],ans ;
int main()
{int n , t = 0 ;while (scanf("%d",&n)!=EOF) {for (int i=1; i<=n; ++i) scanf("%d",a+i);ans = 0 ; for (int i=1; i<=n ; ++i) if (a[i]!=a[i-1]) ++ans ; ans = ans /3 + ( ans % 3>0)  ; ++t ; printf("Case %d: %d\n",t,ans) ; }
}

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B - "Ray, Pass me the dishes!"

线段树，没过，不知道是想错了还是写呲毛了。

#include <iostream>
#include <cstring>
#include <cstdio>using namespace std;/*** 待处理序列A从1到N* min对应的序列是前缀和序列右移一位*/
const int  MaxN = 500000;
#define left  first
#define right second
long long A[MaxN];
struct Range:pair<int,int>{Range(int a=0,int b=0){left  = a;right = b;}long long sum(){return A[right]-A[left-1];}
};
struct node {int l,r,maxPos,minPos;node(int x=0){l=r=maxPos=minPos=x;}long long ans(){return A[l]-A[r-1];}long max(){return A[maxPos];}long min(){return A[minPos-1];}
};
struct SegmentTree
{#define MaxSegmentLength 500000const int root;SegmentTree():root(1){}node data[4*MaxSegmentLength];void newNode(int p,int x){data[p]=node(x);}void maintain(int p){int lch= p*2,rch= p*2+1,t=0;t=(data[lch].ans()>data[rch].ans())?lch:rch;data[p].l=data[t].l;data[p].r=data[t].r;if(data[p].ans()<(data[rch].max()-data[lch].min())){data[p].l=data[lch].minPos;data[p].r=data[rch].maxPos;}data[p].maxPos = (data[lch].max()>data[rch].max())?data[lch].maxPos:data[rch].maxPos;data[p].minPos = (data[lch].min()<data[rch].min())?data[lch].minPos:data[rch].minPos;}void build(int x,int y,int p){if (x==y)newNode(p,x);else {int mid=(x+y)/2;build(x    ,mid,p*2  );build(mid+1,y  ,p*2+1);maintain(p);}}int queryMax(int x,int y,int l,int r,int p){//cout<<"QueryMax  ("<<x<<','<<y<<")  ["<<l<<','<<r<<"] @"<<p<<endl;if (x==l && y==r) return data[p].maxPos;int mid = (l+r)/2,LPos,RPos;//cout<<"        "<<l<<','<<mid<<'|'<<mid+1<<','<<r<<endl;if (y<=mid) return queryMax(x,y,l    ,mid,p*2  );if (x >mid) return queryMax(x,y,mid+1,r  ,p*2+1);LPos=queryMax(x    ,mid,l    ,mid,p*2);RPos=queryMax(mid+1,y  ,mid+1,r  ,p*2+1);//cout<<"     max(L,R)"<<LPos<<' '<<RPos<<endl;return A[LPos]>A[RPos] ? LPos : RPos;}int queryMin(int x,int y,int l,int r,int p){//cout<<"QueryMin  ("<<x<<','<<y<<")  ["<<l<<','<<r<<"] @"<<p<<endl;if (x==l && y==r) return data[p].minPos;int mid = (l+r)/2,LPos,RPos;//cout<<"        "<<l<<','<<mid<<'|'<<mid+1<<','<<r<<endl;if (y<=mid) return queryMin(x,y,l    ,mid,p*2  );if (x >mid) return queryMin(x,y,mid+1,r  ,p*2+1);LPos=queryMin(x    ,mid,l    ,mid,p*2);RPos=queryMin(mid+1,y  ,mid+1,r  ,p*2+1);//cout<<"     min(L,R)"<<LPos<<' '<<RPos<<endl;return A[LPos-1]<A[RPos-1] ? LPos : RPos;}Range query(int x,int y,int l,int r,int p){//cout<<"Query  ("<<x<<','<<y<<")  ["<<l<<','<<r<<"] @"<<p<<endl;if (x==l && y==r) return Range(data[p].l,data[p].r);int mid = (l+r)/2 ;//cout<<"        "<<l<<','<<mid<<'|'<<mid+1<<','<<r<<endl;if (y<=mid) return query(x,y,l    ,mid,p*2  );if (x >mid) return query(x,y,mid+1,r  ,p*2+1);Range LRange,RRange,ans;LRange=query(x    ,mid,l    ,mid,p*2  );RRange=query(mid+1,y  ,mid+1,r  ,p*2+1);ans = LRange.sum()>RRange.sum() ? LRange : RRange;//cout<<"     max(Query)"<<ans.sum()<<endl;int MaxPos=queryMax(mid+1,y  ,mid+1,r  ,p*2+1),MinPos=queryMin(x    ,mid,l    ,mid,p*2  );//cout<<"          MaxPos:"<<MaxPos<<"  "<<"MinPos:"<<MinPos<<endl;if(A[MaxPos]-A[MinPos-1]>ans.sum()){ans.left  = MinPos;ans.right = MaxPos;}//cout<<"      max(max-min)"<<ans.sum()<<endl;return ans;}#undef MaxSegmentLength
};SegmentTree Tree;int main()
{int n,m,CaseNum=0;A[0]=0;while(scanf("%d%d",&n,&m)!=EOF){for(int i=1;i<=n;i++) scanf("%lld",A+i);for(int i=2;i<=n;i++) A[i]+=A[i-1];Tree.build(1,n,Tree.root);printf("Case %d:\n",++CaseNum);for(int i=0;i<m;i++){int a,b;scanf("%d%d",&a,&b);Range ans = Tree.query(a,b,1,n,Tree.root);printf("%d %d\n",ans.left,ans.right);}}/*int N,delta;cin>>N;for(int i=1;i<=N;i++) cin>>A[i];//delta=A[1];//A[1]=A[0]=0;for(int i=2;i<=N;i++)A[i]+=A[i-1];//-delta;//for(int i=1;i<=N;i++)cout<<A[i]<<' ';//cout<<endl;Tree.build(1,N,Tree.root);int a,b;while(cin>>a>>b){Range ans = Tree.query(a,b,1,N,Tree.root);cout<<ans.left<<' '<<ans.right<<endl;//cout<<ans.left<<','<<ans.right<<endl<<endl;}*/return 0;
}
//10
// 1  2  3  4  5  6  7  8  9  10
// 4  5  6  2  9  0 -1 13  9 -10
// 4  9 15 17 26 26 25 38 47  37
// 0  1  2 -2  5 -4 -5  9  5 -14

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C - Plucking fruits

最小瓶颈树，跟kruskal类似

#include<cstdio>
#include<cstring>
#include<algorithm>
const int maxn= 1010 ,maxm = maxn*maxn;struct edge{int u,v,w;};
struct query{int u,v,d,ord;};
int f[maxn],n,m,r;
query q[maxm];
edge e[maxm];
int ord[maxm];
bool ans[maxm];bool compe(edge  a,edge  b){return a.w>b.w ;}
bool compq(query a,query b){return a.d>b.d ;}int Find(int x){ return f[x]==x ? x : f[x]=Find(f[x]);}
int Union(int x,int y){int fx = Find(x),fy = Find(y);if(fx!=fy){f[fx]=fy;}return 0;
} int main()
{int caseNuym = 1;while(scanf("%d%d%d",&n,&m,&r)!=EOF) {for(int i=1;i<=n;i++)f[i]=i;memset(ans,0,sizeof(ans));for(int i=0;i<m;i++)scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);for(int i=0;i<r;i++){scanf("%d%d%d",&q[i].u,&q[i].v,&q[i].d);q[i].ord = i;}std::sort(e,e+m,compe);std::sort(q,q+r,compq);int head = 0;for(int i=0;i<r;i++){for(;(e[head].w>=q[i].d)&&(head<m);head++)Union(e[head].u,e[head].v);if(Find(q[i].u)==Find(q[i].v))ans[q[i].ord] = 1;else ans[q[i].ord] = 0;}printf("Case %d:\n",caseNuym ); for (int i=0;i<r;++i)if (ans[i]) puts("yes"); else puts("no"); ++caseNuym ; }}

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F - Remember the Word

动态规划f[i]表示单词的i前缀串划分为集合中字符的方案数

转移方程 f[i]=sigma(f[i-s[j].len]) s[j]是word[0..i]一个后缀

边界 f[0]=1

实际上写的时候考虑的是向后更新会比较方便。

f[i+s[j].len]+=f[i] (s[j]是word[i..end]一个前缀）

利用字典树找到一个状态的所有后继。

#include <cstdio>
#include <cstring>
#include <queue>const long long maxnode    = 4010*100;
const long long SIGMA_SIZE = 26;
const long long maxlen     = 100;
const long long maxn       = 300010;
const long long MOD        = 20071027;using namespace std;char s[maxn];
int f[maxn];struct Trie
{int ch[maxnode][SIGMA_SIZE];int val[maxnode];int size;Trie(){clear();}void clear(){size=1;memset(ch[0],0,sizeof(ch[0]));memset(val,0,sizeof(val));}int index(char t){return t-'a';};int buildNewNode(int u){memset(ch[size],0,sizeof(ch[size]));val[size]=u;return size++;}void insert(char *s){int u = 0;for(;*s;s++){int c = index(*s);//忘记调用idx()WA了好几次if(!ch[u][c])ch[u][c]=buildNewNode(0);u=ch[u][c];}val[u]=1;//val[u]=v
    }void find(int i,char *s){int u=0,len=0;for(;*s;s++){int c = index(*s);//忘记调用idx()WA了好几次if(ch[u][c]){len++;u=ch[u][c];f[i+len]=(val[u]*f[i]+f[i+len])%MOD;//if(val[u]){find an end }
            }elsebreak;}}
};Trie trie;int main()
{int CaseNum = 0;while(scanf("%s",s)!=EOF){int n,len=strlen(s);memset(f,0,sizeof(f));f[0]=1;trie.clear();scanf("%d",&n);for(int i=0;i<n;i++){char key[110];scanf("%s",key);trie.insert(key);}for(int i=1;i<=len;i++)trie.find(i-1,s+i-1);printf("Case %d: %d\n",++CaseNum,f[len]);}return 0;
}

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I - Cake slicing

4D/1D的动态规划，就是枚举这一刀是横切还是竖切，记忆化搜索。

#include <iostream>
#include <cstring>
#include <cstdio>using namespace std;const int maxn   = 20+10;
const int maxInt = 0x3f3f3f3f;int A[maxn][maxn],f[maxn][maxn][maxn][maxn];int sum(int x,int y,int a,int b){return A[a][b]-A[x-1][b]-A[a][y-1]+A[x-1][y-1];}int dfs(int x,int y,int a,int b){int S = sum(x,y,a,b),&ans=f[x][y][a][b];if(S==0)return maxInt;if(S==1)return 0;if(ans) return ans;ans=maxInt;for (int i = x; i < a; ++i){ans = min(dfs(x,y,i,b)+dfs(i+1,y,a,b)+b-y+1,ans);}for (int i = x; i < a; ++i){ans = min(dfs(x,y,a,i)+dfs(x,i+1,a,b)+a-x+1,ans);}return ans;
}int main(int argc, char const *argv[])
{int n,m,k,CaseNum=0;while(cin>>n>>m>>k){memset(A,0,sizeof(A));memset(f,0,sizeof(f));for(int i=0;i<k;i++){int x,y;cin>>x>>y;A[x][y]=1;}for (int i = 1; i <= n; ++i){for (int j = 1; j <= m; ++j){A[i][j]+=A[i-1][j]+A[i][j-1]-A[i-1][j-1];}}int ans = dfs(1,1,n,n);cout<<"Case "<<++CaseNum<<": "<<ans<<endl;}return 0;
}