u，v，w。

这场考过。

---

T1 u

差分裸题

#include<bits/stdc++.h>
using namespace std;
const int N=5000;
int n,m;
long long a[N][N],b[N][N],f[N][N];
long long ans=0;
int _max(int a,int b)
{
    return a>b?a:b;
}
int _min(int a,int b)
{
    return a<b?a:b;
}
int read()
{
    int s=0,w=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
    while('0'<=ch&&ch<='9'){s=(s<<1)+(s<<3)+ch-'0';ch=getchar();}
    return s*w;
}
int main()
{
    n=read();m=read();
    for(int i=1,r,c,l,s;i<=m;i++)
    {
        r=read();
        c=read();
        l=read();
        s=read();
        a[r][c]+=s;
        a[r+l][c+l]-=s;
        b[r+l][c]-=s;
        b[r+l][l+c]+=s;
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            b[i][j+1]+=b[i][j];
            a[i][j]+=a[i-1][j-1];
        }
    }
    for(int j=1;j<=n;j++)
    {
        for(int i=1;i<=n;i++)
        {
            a[i][j]+=a[i-1][j];
            b[i][j]+=b[i-1][j];
        }
    }
    for(int j=1;j<=n;j++)
    {
        for(int i=1;i<=n;i++)
        {
            f[i][j]+=a[i][j]+b[i][j];
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            ans^=f[i][j];
        }
    }
    printf("%lld\n",ans);
    return 0;
}

---

T2 状压DP+分段记忆化（数组+map）

//不是我代码，转自网络，侵删
#include <bits/stdc++.h>
using namespace std;
map <int,double > mp;
int n,k;
char s[100];
double f[1<<25];
 
double dfs(int node,int tot)
{
    if(tot== n - k) 
        return 0;
    if(tot > 24 && mp.find(node)!= mp.end()) 
        return mp[node];
    if(tot <= 24 && f[node]!= -1)
        return f[node];
    double sum= 0;
    for(int i=1;i<= (tot >>1);i++) 
    {
        int color1= (node >> (i-1)) &1;
        int color2= (node >> (tot - i)) &1;
        int node1= (node >>1) & (~ ((1 << (i-1)) -1)) | (node & ((1 << (i-1)) -1));
        int node2= (node >>1) & (~ ((1 << (tot - i)) -1)) | (node & ((1 << (tot - i)) -1));
        sum += 2 * max(dfs(node1,tot -1) + color1,dfs(node2,tot -1) + color2) / tot;
    }
    if(tot &1)
     {
        int i= (tot +1) >>1; 
        int color= (node >> (i-1)) &1;
        int st= (node >>1) & (~ ((1 << (i-1)) -1)) | (node & ((1 << (i-1)) -1));
        sum += (dfs(st,tot -1) + color) / tot;
    }
    return (tot > 24)?mp[node]= sum:f[node]= sum;
}

int main()
{
    scanf("%d %d %s",&n,&k,s);
    int node= 0;
    for(int i= 0;i< (1 << 25);i++)
        f[i]= -1;
    for(int i=1;i<= n;i++)
        node |= (s[i-1]== 'W') << (n - i);
    node |= (1 << n);
    printf("%.8f\n",dfs(node,n));
    return 0;
}

---

T3 w

树形DP，f[u][0/1]（二元组）表示u点上面的边反转/不反转的最少路径和最短路。

#include<bits/stdc++.h>
using namespace std;
const int inf=0x3f3f3f3f,N=300000;
struct bian
{
    int nxt,to,w;
}
b[N];
int n,head[N],cnt;
pair<int,int> f[2][N];
void add(int x,int y,int w)
{
    b[++cnt].nxt=head[x];
    b[cnt].to=y;
    b[cnt].w=w;
    head[x]=cnt;
}
pair<int,int> pls(pair<int,int> a,pair<int,int> b)
{
    return make_pair(a.first+b.first,a.second+b.second);
}
pair<int,int> min(pair<int,int> x,pair<int,int> y)
{
    if(x.first<y.first)
        return x;
    else if(x.first>y.first)
        return y;
    else if(x.second<y.second)
        return x;
    else
         return y;
}
void dfs(int x,int fa,int w)
{
    pair<int,int> w1=make_pair(inf,inf);
    pair<int,int> w2=make_pair(0,0);
    for(int i=head[x];i;i=b[i].nxt)
    {
        if(b[i].to!=fa)
        {
            dfs(b[i].to,x,b[i].w);
            pair<int,int> flag1=min(pls(w1,f[0][b[i].to]),pls(w2,f[1][b[i].to]));
            pair<int,int> flag2=min(pls(w1,f[1][b[i].to]),pls(w2,f[0][b[i].to]));
            w1=flag1;
            w2=flag2;
        }
    }
    if(w==1)
        f[0][x]=make_pair(inf,inf);
    else
        f[0][x]=min(make_pair(w1.first+1,w1.second),w2);
    if(w==0)
        f[1][x]=make_pair(inf,inf);
    else
        f[1][x]=min(make_pair(w1.first,w1.second+1),make_pair(w2.first+1,w2.second+1));
}
int main()
{
    scanf("%d",&n);
    for(int i=1,from,to,c,d;i<n;i++)
    {
        scanf("%d%d%d%d",&from,&to,&c,&d);
        if(d==2)
        {
            add(from,to,2);
            add(to,from,2);
        }
        else
        {
            add(from,to,c!=d);
            add(to,from,c!=d);
        }
    }
    dfs(1,0,2);
    printf("%d %d\n",f[0][1].first>>1,f[0][1].second);
    return 0;
}