HDU5905（树dp)

Black White Tree

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 80 Accepted Submission(s): 40

Problem Description
Alex has a tree T with n vertices conveniently labeled with 1,2,…,n. Each vertex of the tree is colored black or white. For every integer pair (a,b), Alex wants to know whether there is a subtree of T with exact a white vertices and b black vertices.

A subtree of T is a subgraph of T that is also a tree.

Input
There are multiple test cases. The first line of input contains an integer T (1≤T≤100) indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤2000). The next line contains a binary string s length n where si denoting the color of the i-th vertex (‘0’ means white, while ‘1’ means black). Each of the following n−1 lines contains two integers ai,bi, denoting an edge between vertices ai and bi (1≤ai,bi≤n).

Output
For each test case, output an integer W=∑na=0∑nb=0(a+1)(b+1)S(a,b). S(a,b)=1 if there is a subtree with exact a white vertices and b black vertices, or 0 otherwise.

Sample Input
3
4
1010
1 4
2 4
3 4
3
101
1 2
2 3
10
1010111001
1 2
1 3
1 8
2 4
2 6
2 7
3 9
4 5
7 10

Sample Output
33
15
365

题意：这是bc题，有中文题，自己去找。

解题思路：有一个结论，就是大小为i的子树中，肯定有一个里面的黑点数最大，我们设为Max，肯定有一个里面的黑点数最小，我们设为Min，那么这个结论就是这个树一定存在大小为i,黑点数介于Max与Min之间的子树，具体怎么证明，本弱鸡也不会，跪请大佬解释下怎么证明，那么知道了这个结论后就好办了，我们只需要求出大小为i的子图中Max与Min就行。具体做发看代码注释。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e3 + 10;
const ll inf = 1e9;
ll Max[maxn];//Max[i]表示大小为i的连通图黑点数的最大值
ll Min[maxn];//Min[i]表示大小为i的连通图黑点数的最小值
ll dpMax[maxn][maxn];//dpMax[i][j]表示以i为根的子树大小为j的最大黑点数
ll dpMin[maxn][maxn];//dpMin[i][j]表示以i为根的子树大小为j的最小黑点数
ll dp[maxn];//dp[i]表示以i为根的子树的大小
bool visit[maxn];
int n;
char s[maxn];
vector<int> g[maxn];
void init()
{
    memset(visit,false,sizeof(visit));
    memset(dp,0,sizeof(dp));
    for(int i = 1; i <= n; i++)
    {
        g[i].clear();
        Max[i] = -inf;
        Min[i] = inf;
    }
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            dpMax[i][j] = -inf;
            dpMin[i][j] = inf;
        }
    }

}
void dfs(int root)
{
    if(visit[root]) return;
    visit[root] = true;
    dp[root] = 1;
    for(int i = 0; i < g[root].size(); i++)
    {
        int v = g[root][i];
        if(!visit[v])
        {
            dfs(v);
            dp[root] += dp[v];
        }
    }
}
void solve(int root)
{
    if(visit[root]) return;
    dpMax[root][0] = dpMin[root][0] = 0;
    if(s[root] == '1')
    {
        dpMax[root][1] = dpMin[root][1] = 1;
    }
    else
    {
        dpMax[root][1] = dpMin[root][1] = 0;
    }
    visit[root] = true;
    int ans = 1;
    for(int i = 0; i < g[root].size(); i++)
    {
        int v = g[root][i];
        if(!visit[v])
        {
            solve(v);
            for(int j = ans; j > 0; j--)
            {
                for(int k = 1; k <= dp[v]; k++)
                {
                    dpMax[root][j + k] = max(dpMax[root][j + k],dpMax[root][j] + dpMax[v][k]);
                    dpMin[root][j + k] = min(dpMin[root][j + k],dpMin[root][j] + dpMin[v][k]);
                }
            }
            ans += dp[v];
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        init();
        scanf("%s",s + 1);
        int u,v;
        for(int i = 1; i < n; i++)
        {
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        dfs(1);//假设以一为根
        memset(visit,false,sizeof(visit));
        solve(1);
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                Max[j] = max(Max[j],dpMax[i][j]);
                Min[j] = min(Min[j],dpMin[i][j]);
            }
        }
        ll sum = 1;//a = 0,b = 0的特殊情况
        for(int i = 1; i <= n; i++)
        {
            for(int j = Max[i]; j >= Min[i]; j--)
            {
                sum += (j + 1)*(i - j + 1);
            }
        }
        printf("%lld\n",sum);

    }
    return 0;
}