dp 杂练/专练

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最近各种各样的事让我感受到了自己dp是多么弱，于是开始写dp
bzoj1055
dp[i][j][k]表示i到j能否缩成k
没了

//Copyright(c)2016 liuchenrui#include<bits/stdc++.h>#define N 255using namespace std;char p[]={'W','I','N','G'},s[N],a[4][20][3];int q[N],t[4],dp[N][N][4];bool dodp(int l,int r,int k){    if(l==r)return s[l]==p[k];    int &res=dp[l][r][k];    if(res!=-1)return res;    for(int i=1;i<=t[k];i++)        for(int j=l;j<=r-1;j++)            if(dodp(l,j,q[a[k][i][0]])&&dodp(j+1,r,q[a[k][i][1]]))                return res=1;    return res=0;}int main(){    freopen("xxx.in","r",stdin);    freopen("xxx.out","w",stdout);    memset(dp,-1,sizeof(dp));    q['W']=0;q['I']=1;q['N']=2;q['G']=3;    for(int i=0;i<4;i++){        cin>>t[i];    }    for(int i=0;i<4;i++){        for(int j=1;j<=t[i];j++){            scanf("%s",a[i][j]);        }    }    scanf("%s",s+1);int n=strlen(s+1);    bool flag=0;    for(int i=0;i<4;i++){        if(dodp(1,n,i)){            flag=1,printf("%c",p[i]);        }    }    if(!flag)puts("The name is wrong!");}

bzoj1090
dp[i][j]表示i到j的最小距离
n3枚举，暴力判断即可

//Copyright(c)2016 liuchenrui#include<bits/stdc++.h>#define inf 1000000000#define N 105using namespace std;int dp[N][N],n;char s[N];int get(int x){    if(x<10)return 1;    if(x==100)return 3;    return 2;}int calc(int l1,int r1,int l2,int r2){    if((r2-l2+1)%(r1-l1+1))return inf;    for(int i=l2,j=l1;i<=r2;i++,j++){        if(j==r1+1)j=l1;        if(s[i]!=s[j])return inf;    }    return dp[l1][r1]+2+get((r2-l2+1)/(r1-l1+1)+1);}int main(){    freopen("xxx.in","r",stdin);    freopen("xxx.out","w",stdout);    scanf("%s",s+1);n=strlen(s+1);    memset(dp,63,sizeof dp);    for(int i=1;i<=n;i++){        dp[i][i]=1;    }    for(int i=1;i<=n;i++){        for(int j=1;j<=n-i+1;j++){            int l=j,r=i+j-1;            for(int k=l;k<r;k++){                dp[l][r]=min(dp[l][r],dp[l][k]+dp[k+1][r]);                dp[l][r]=min(dp[l][r],calc(l,k,k+1,r));            }        }    }    cout<<dp[1][n]<<endl;}

bzoj1334
翻译人/出题人语死早
题意：给一个集合，选出一个子集a，使得这个子集和sum(a)比全部的和的一半sum(tot)大，并且对于任意子集内的元素a[i]，sum(a)−a[i]比sum(tot)小
这么水的背包？
可以压位，懒得写了

//Copyright(c)2016 liuchenrui#include<bits/stdc++.h>using namespace std;inline void splay(int &v){    v=0;char c=0;int p=1;    while(c<'0' || c>'9'){if(c=='-')p=-1;c=getchar();}    while(c>='0' && c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();}    v*=p;}int dp[100005],a[305],sum,ans,n;int main(){    //freopen("xxx.in","r",stdin);    //freopen("xxx.out","w",stdout);    splay(n);    for(int i=1;i<=n;i++)splay(a[i]),sum+=a[i];    sort(a+1,a+n+1);dp[0]=1;    for(int i=n;i>=1;i--){        for(int j=sum/2+a[i];j>=a[i];j--){            if(dp[j-a[i]]){                dp[j]=1;ans=max(j,ans);            }        }    }    printf("%d",ans);}

bzoj3687
interesting
dp[i]表示i能否达到
相同的数xor会抵消
看代码要用心体会

//Copyright(c)2016 liuchenrui#include<bits/stdc++.h>using namespace std;inline void splay(int &v){    v=0;char c=0;int p=1;    while(c<'0' || c>'9'){if(c=='-')p=-1;c=getchar();}    while(c>='0' && c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();}    v*=p;}bitset<2000001>a;int main(){    int n;splay(n);int x,ans=0;a[0]=1;    for(;n--;)scanf("%d",&x),a^=a<<x;    for(int i=1;i<=2000000;i++)if(a[i])ans^=i;    cout<<ans<<endl;}

bzoj3573
这道题被抓去直播写代码了QAQ
一个权值确定会带领全图的权值确定
令p[1]=0,p[x]=(这个点在1点为1的情况下权值为多少)
然后每个点反过来取log
然后排序，记录相同的权值，答案为n−最多的权值个数

//Copyright(c)2016 liuchenrui#include<bits/stdc++.h>using namespace std;inline void splay(int &v){    v=0;char c=0;int p=1;    while(c<'0' || c>'9'){if(c=='-')p=-1;c=getchar();}    while(c>='0' && c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();}    v*=p;}#define N 500010int nxt[N*2],fir[N],to[N*2],a[N],sz,d[N],n;double p[N];void add(int x,int y){    nxt[++sz]=fir[x],fir[x]=sz,to[sz]=y;}void dfs(int x,int fa){    for(int u=fir[x];u;u=nxt[u]){        if(to[u]!=fa){            p[to[u]]=p[x]+log(double(d[x]));            dfs(to[u],x);        }    }}int main(){    freopen("xxx.in","r",stdin);    freopen("xxx.out","w",stdout);    splay(n);    for(int i=1;i<=n;i++){        splay(a[i]);        }    for(int i=1;i<n;i++){        int x,y;splay(x),splay(y);        add(x,y),add(y,x);        d[x]++,d[y]++;    }    for(int i=2;i<=n;i++)d[i]--;    p[1]=0;dfs(1,0);    for(int i=1;i<=n;i++){        p[i]+=log(double(a[i]));    }    sort(p+1,p+n+1);    int now=1,ans=0;    for(int i=1;i<=n;i++){        if(p[i]-p[now]>1e-5){            ans=max(ans,i-now);            now=i;        }    }    ans=max(ans,n-now+1);    cout<<n-ans<<endl; }

bzoj1369
据clj讲，这题颜色不会超过10，我开到5就够了
dp[i][j]表示i点，j颜色最小代价
剩下自己yy就能做出来辣

//Copyright(c)2016 liuchenrui#include<bits/stdc++.h>#define N 100010using namespace std;inline void splay(int &v){    v=0;char c=0;int p=1;    while(c<'0' || c>'9'){if(c=='-')p=-1;c=getchar();}    while(c>='0' && c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();}    v*=p;}int dp[N][6],n,fir[N],nxt[N*2],to[N*2],sz;void add(int x,int y){    nxt[++sz]=fir[x],fir[x]=sz,to[sz]=y;}void dfs(int x,int fa){    for(int u=fir[x];u;u=nxt[u]){        if(to[u]==fa)continue;        dfs(to[u],x);    }    for(int i=1;i<=5;i++){        int cnt=i;        for(int u=fir[x];u;u=nxt[u]){            if(to[u]==fa)continue;            int mi=1<<28;            for(int j=1;j<=5;j++){                if(i!=j){                    mi=min(mi,dp[to[u]][j]);                }            }            cnt+=mi;        }        dp[x][i]=cnt;    }}int main(){    freopen("xxx.in","r",stdin);    freopen("xxx.out","w",stdout);    splay(n);    for(int i=1,x,y;i<n;i++){        splay(x),splay(y);        add(x,y),add(y,x);    }    dfs(1,0);    int ans=1<<28;    for(int i=1;i<=5;i++){        ans=min(ans,dp[1][i]);    }    cout<<ans<<endl;}

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