【DP】树形DP题目总结
最近在研究树形DP，所以做了几道比较典型的树形DP的题目。大多数都是一些比较经典的题，大部分都用了多叉转二叉，题目也比较类似。感觉树形DP的模式都比较固定，基本都是先建树，然后dfs，方程也基本都是枚举左子树和右子树各自分配资源得到根的最大值，具体见题目吧。
1、rqnoj30：愚蠢的矿工
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1000 + 10;
const int maxm = 100 + 10;
struct tree
{
int l,r;
tree()
{
l = r = -1;
}
}Tree[maxn];
int n,m;
int w[maxn];
int f[maxn][maxm];
bool map[maxn][maxn];
void init()
{
freopen("rqnoj30.in","r",stdin);
freopen("rqnoj30.out","w",stdout);
}
void readdata()
{
memset(map,false,sizeof(map));
scanf("%d%d",&n,&m);
for(int i = 1;i <= n;i++)
{
scanf("%d",&w[i]);
}
for(int i = 1;i <= n;i++)
{
int a,b;
scanf("%d%d",&a,&b);
map[a][b] = true;
}
}
void maketree(int k)
{
for(int i = 1;i <= n;i++)
{
if(map[k][i])
{
if(Tree[k].l == -1)
{
Tree[k].l = i;
}
else
{
int t = Tree[k].l;
while(Tree[t].r != -1)t = Tree[t].r;
Tree[t].r = i;
}
}
}
if(Tree[k].l != -1)maketree(Tree[k].l);
if(Tree[k].r != -1)maketree(Tree[k].r);
}
int max(int a,int b)
{
return a > b ? a : b;
}
int dfs(int d,int k)
{
if(d == -1)return 0;
if(f[d][k] != -1)return f[d][k];
f[d][k] = dfs(Tree[d].r,k);
for(int i = 0;i <= k - 1;i++)
{
f[d][k] = max(dfs(Tree[d].l,i) + dfs(Tree[d].r,k-i-1) + w[d],f[d][k]);
}
return f[d][k];
}
void solve()
{
memset(f,-1,sizeof(f));
for(int i = 1;i <= n;i++)f[i][0] = 0;
maketree(0);
printf("%d",dfs(0,m+1));
}
int main()
{
init();
readdata();
solve();
return 0;
}
2、ural1018：二叉苹果树
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 100 + 10;
struct tnode
{
int l,r;
}tree[maxn];
int map[maxn][maxn];
int w[maxn];
int f[maxn][maxn];
int n,q;
void init()
{
freopen("ural1018.in","r",stdin);
freopen("ural1018.out","w",stdout);
}
void readdata()
{
memset(map,-1,sizeof(map));
scanf("%d%d",&n,&q);
for(int i = 1;i < n;i++)
{
int l,r,w;
scanf("%d%d%d",&l,&r,&w);
map[l][r] = map[r][l] = w;
}
}
void maketree(int v)
{
for(int i = 1;i <= n;i++)
{
if(map[v][i] >= 0)
{
tree[v].l = i;
w[i] = map[v][i];
map[v][i] = map[i][v] = -1;
maketree(i);
break;
}
}
for(int i = 1;i <= n;i++)
{
if(map[v][i] >= 0)
{
tree[v].r = i;
w[i] = map[v][i];
map[v][i] = map[i][v] = -1;
maketree(i);
break;
}
}
}
int max(int a,int b)
{
return a > b ? a : b;
}
void dfs(int v,int k)
{
if(k == 0)
f[v][k] = 0;
else if(tree[v].l == 0 && tree[v].r == 0)
{
f[v][k] = w[v];
}
else
{
f[v][k] = 0;
for(int i = 0;i < k;i++)
{
if(f[tree[v].l][i] == 0)dfs(tree[v].l,i);
if(f[tree[v].r][k-i-1] == 0)dfs(tree[v].r,k-i-1);
f[v][k] = max(f[v][k],f[tree[v].l][i]+f[tree[v].r][k-i-1] + w[v]);
}
}
}
void solve()
{
maketree(1);
dfs(1,q+1);
int ans = f[1][q+1];
printf("%d",ans);
}
int main()
{
init();
readdata();
solve();
return 0;
}
3、tyvj1051：选课
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 300 + 10;
struct tree
{
int l,r;
tree()
{
l = r = -1;
}
}Tree[maxn];
bool map[maxn][maxn];
int num[maxn];
int f[maxn][maxn];
int n,m;
void init()
{
freopen("tyvj1051.in","r",stdin);
freopen("tyvj1051.out","w",stdout);
}
void readdata()
{
memset(map,false,sizeof(map));
scanf("%d%d",&n,&m);
for(int i = 1;i <= n;i++)
{
int t,w;
scanf("%d%d",&t,&w);
map[t][i] = true;
num[i] = w;
}
}
void maketree(int k)
{
for(int i = 1;i <= n;i++)
{
if(map[k][i])
{
if(Tree[k].l == -1)
{
Tree[k].l = i;
}
else
{
int t = Tree[k].l;
while(Tree[t].r != -1)t = Tree[t].r;
Tree[t].r = i;
}
}
}
if(Tree[k].l != -1)maketree(Tree[k].l);
if(Tree[k].r != -1)maketree(Tree[k].r);
}
int max(int a,int b)
{
return a > b ? a : b;
}
int dfs(int t,int m)
{
if(t == -1)return 0;
if(f[t][m] != -1)return f[t][m];
f[t][m] = dfs(Tree[t].r,m);
for(int i = 0;i <= m - 1;i++)
{
f[t][m] = max(dfs(Tree[t].l,i) + dfs(Tree[t].r,m-i-1) + num[t],f[t][m]);
}
return f[t][m];
}
void solve()
{
memset(f,-1,sizeof(f));
for(int i = 1;i <= n;i++)f[i][0] = 0;
maketree(0);
printf("%d",dfs(0,m+1));
}
int main()
{
init();
readdata();
solve();
return 0;
}
4、tyvj1052：没有上司的舞会
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 6000 + 10;
struct tree
{
int l,r;
tree()
{
l = r = -1;
}
}Tree[maxn];
int root,n;
int f[maxn][2];
int w[maxn];
bool map[maxn][maxn];
bool flag[maxn];
void init()
{
freopen("tyvj1052.in","r",stdin);
freopen("tyvj1052.out","w",stdout);
}
void readdata()
{
memset(flag,true,sizeof(flag));
memset(map,false,sizeof(map));
scanf("%d",&n);
for(int i = 1;i <= n;i++)
{
scanf("%d",&w[i]);
}
for(int i = 1;i < n;i++)
{
int l,k;
scanf("%d%d",&l,&k);
map[k][l] = true;
flag[l] = false;
}
for(int i = 1;i <= n;i++)
{
if(flag[i])
{
root = i;
break;
}
}
}
void maketree(int k)
{
for(int i = 1;i <= n;i++)
{
if(map[k][i])
{
if(Tree[k].l == -1)
{
Tree[k].l = i;
}
else
{
int t = Tree[k].l;
while(Tree[t].r != -1)t = Tree[t].r;
Tree[t].r = i;
}
}
}
if(Tree[k].l != -1)maketree(Tree[k].l);
if(Tree[k].r != -1)maketree(Tree[k].r);
}
int max(int a,int b)
{
return a > b ? a : b;
}
int dfs(int k,int s)
{
if(f[k][s] != -1)return f[k][s];
f[k][s] = 0;
if(s == 1)f[k][s] = w[k];
for(int i = Tree[k].l;i != -1;i = Tree[i].r)
{
int com;
if(s == 0)
{
com = max(dfs(i,0),dfs(i,1));
if(com >= 0)f[k][s] += com;
}
if(s == 1)
{
com = dfs(i,0);
if(com >= 0)f[k][s] += com;
}
}
return f[k][s];
}
void solve()
{
memset(f,-1,sizeof(f));
maketree(root);
int ans = max(dfs(root,0),dfs(root,1));
printf("%d",ans);
}
int main()
{
init();
readdata();
solve();
return 0;
}
5、tyvj1513：小胖守皇宫
这道题题意有点不好理解，每个节点有三种状态：第一种是在该节点放士兵；第二种是在该节点不放，但在它的儿子节点放了士兵；第三种是该节点不放，它被父亲节点影响。所以第一种情况可以由三种情况转移，第二种情况能由1,2种情况转移，第三种情况只能由第二种情况来转移。
#include<cstdio>
#include<cstring>
using namespace std;
const int inf = 0x3ffff;
const int maxn = 1500 + 10;
const int maxm = 5;
struct tree
{
int l,r;
tree()
{
l = r = -1;
}
}Tree[maxn];
int w[maxn];
int f[maxn][maxm];
int n,root;
bool vis[maxn][maxm];
bool map[maxn][maxn];
bool flag[maxn];
void init()
{
freopen("tyvj1513.in","r",stdin);
freopen("tyvj1513.out","w",stdout);
}
void readdata()
{
memset(flag,false,sizeof(flag));
memset(map,false,sizeof(map));
scanf("%d",&n);
for(int i = 1;i <= n;i++)
{
int t1,m;
scanf("%d",&t1);
scanf("%d",&w[t1]);
scanf("%d",&m);
for(int j = 1;j <= m;j++)
{
int t2;
scanf("%d",&t2);
map[t1][t2] = true;
flag[t2] = true;
}
}
for(int i = 1;i <= n;i++)
{
if(flag[i] == false)
{
root = i;
break;
}
}
}
void maketree(int k)
{
for(int i = 1;i <= n;i++)
{
if(map[k][i])
{
if(Tree[k].l == -1)
{
Tree[k].l = i;
}
else
{
int t = Tree[k].l;
while(Tree[t].r != -1)t = Tree[t].r;
Tree[t].r = i;
}
}
}
if(Tree[k].l != -1)maketree(Tree[k].l);
if(Tree[k].r != -1)maketree(Tree[k].r);
}
int min(int a,int b)
{
return a < b ? a : b;
}
int dfs(int k,int s)
{
if(vis[k][s])return f[k][s];
vis[k][s] = true;
int sum = 0,temp = inf;
for(int i = Tree[k].l;i != -1;i = Tree[i].r)
{
if(s == 1)
{
f[k][s] += min(dfs(i,1),min(dfs(i,2),dfs(i,3)));
}
if(s == 2)
{
f[k][s] += min(dfs(i,1),dfs(i,2));
if(f[i][1] - min(f[i][1],f[i][2]) < temp)
{
temp = f[i][1] - min(f[i][1],f[i][2]);
}
}
if(s == 3)
{
f[k][s] += dfs(i,2);
}
}
if(s == 1)f[k][s] += w[k];
if(s == 2)f[k][s] += temp;
return f[k][s];
}
void solve()
{
memset(vis,false,sizeof(vis));
memset(f,0,sizeof(f));
maketree(root);
int ans = min(dfs(root,1),dfs(root,2));
printf("%d",ans);
}
int main()
{
init();
readdata();
solve();
return 0;
}
The End...