HDU 3534 Tree 树形dp统计

题意：给定n（范围不明确，不过10000可以过）个点的树，问树的直径有多少条。
题解：维护一个子根节点到子树中叶子节点的最长路，次长路和对应的个数，注意路之间不能在同一棵子树内，然后统计每棵子树的直径，最后遍历得到答案。

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#include <iostream>#include <cstdio>#include <memory.h>using namespace std;const int inf = 1 << 29;const int maxn = 10002;struct node{    int v,w;    int next;}edge[maxn << 1];struct A{    int path,num;    int dep1,dep2;    int num1,num2;}ans[maxn];int head[maxn];int n,idx;void init(){    memset(head,-1,sizeof(head));    idx = 0;    return;}void addedge(int u,int v,int w){    edge[idx].v = v;    edge[idx].w = w;    edge[idx].next = head[u];    head[u] = idx++;    edge[idx].v = u;    edge[idx].w = w;    edge[idx].next = head[v];    head[v] = idx++;    return;}void read(){    int u,v,w;    for(int i=1;i<n;i++)    {        scanf("%d %d %d",&u,&v,&w);        addedge(u,v,w);    }    return;}void dfs(int st,int pre){    ans[st].dep1 = ans[st].dep2 = -inf;    ans[st].num1 = ans[st].num2 = ans[st].num = 0;    bool leaf = true;    for(int i=head[st];i != -1;i=edge[i].next)    {        if(edge[i].v == pre) continue;        dfs(edge[i].v , st);        leaf = false;        int tmp = ans[edge[i].v].dep1 + edge[i].w;        if(tmp > ans[st].dep1)        {            ans[st].dep2 = ans[st].dep1;            ans[st].num2 = ans[st].num1;            ans[st].dep1 = tmp;            ans[st].num1 = ans[edge[i].v].num1;        }        else if(tmp == ans[st].dep1)        {            ans[st].num1 += ans[edge[i].v].num1;        }        else if(tmp > ans[st].dep2)        {            ans[st].dep2 = tmp;            ans[st].num2 = ans[edge[i].v].num1;        }        else if(tmp == ans[st].dep2)        {            ans[st].num2 += ans[edge[i].v].num1;        }    }    if(leaf)    {        ans[st].dep1 = ans[st].path = 0;        ans[st].num1 = ans[st].num = 1;        return;    }    int c1 = 0,c2 = 0;    for(int i=head[st];i != -1;i=edge[i].next)    {        if(edge[i].v == pre) continue;        int tmp = ans[edge[i].v].dep1 + edge[i].w;        if(tmp == ans[st].dep1) c1++;        else if(tmp == ans[st].dep2) c2++;    }    int type;   //当前子树内的直径有三种情况，最大+最大（dep1+dep1），最大+次大（dep1+dep2），最大（dep1）    if(c1 > 1) type = 1;    else if(c2 > 0) type = 2;    else type = 3;    if(type == 1)    {        ans[st].path = ans[st].dep1 * 2;        int sum = 0;        for(int i=head[st];i != -1;i=edge[i].next)        {            if(edge[i].v == pre) continue;            if(ans[edge[i].v].dep1 + edge[i].w == ans[st].dep1)            {                ans[st].num += sum * ans[edge[i].v].num1;                sum += ans[edge[i].v].num1;            }        }    }    else if(type == 2)    {        ans[st].path = ans[st].dep1 + ans[st].dep2;        for(int i=head[st];i != -1;i=edge[i].next)        {            if(edge[i].v == pre) continue;            if(ans[edge[i].v].dep1 + edge[i].w == ans[st].dep2)            {                ans[st].num += ans[st].num1 * ans[edge[i].v].num1;            }        }    }    else    {        ans[st].path = ans[st].dep1;        ans[st].num = ans[st].num1;    }    return;}void solve(){    dfs(1,0);    int res = -inf,num = 0;    for(int i=1;i<=n;i++)    {        if(ans[i].path > res)        {            res = ans[i].path;            num = ans[i].num;        }        else if(ans[i].path == res)        {            num += ans[i].num;        }    }    printf("%d %d\n",res,num);    return;}int main(){    while(~scanf("%d",&n))    {        init();        read();        solve();    }    return 0;}