hdu 4003 Find Metal Mineral

#include <cstdio>#include <cstring>int n, s, k;const int MAXN = 10010;struct EDGE{    int v, next, w;}edge[MAXN * 2];int head[MAXN];int cnt;int dp[MAXN][11];void addedge(int u, int v, int val){    edge[cnt].v = v;    edge[cnt].next = head[u];    edge[cnt].w = val;    head[u] = cnt++;}void dfs(int t, int fa){    //dp[i][j] 暂时在i的子树中最多用j个机器人达到的最优值（J个机器人回去I节点）    for(int p = head[t]; p != -1; p = edge[p].next)    {        if(edge[p].v == fa)  continue;        dfs(edge[p].v, t);    }    if(head[t] == -1 || (edge[head[t]].next == -1 && fa != -1))   //一颗空树或者叶子节点，返回    {        for(int i = 0; i <= k; i++)//注意一下这个到底是否成立        {            dp[t][i] = 0;        }        return;    }    for(int p = head[t]; p != -1; p = edge[p].next)    {        int v = edge[p].v;        if(v == fa)  continue;//!!!!        for(int j = k; j >= 0; j--)        {            if(dp[t][j] == -1)//j个不返回            {                if(j > 0)                {                    dp[t][j] = dp[v][0] + 2 * edge[p].w;                    for(int i = 1; i <= j; i++)                    if(dp[t][j] > dp[v][i] + i * edge[p].w)                    {                         dp[t][j] = dp[v][i] + i * edge[p].w;                    }                }                else                {                    dp[t][j] = dp[v][j] + 2 * edge[p].w;                }                continue;            }            else            {                dp[t][j] += dp[v][0] + 2 * edge[p].w;                for(int i = 1; i <= j; i++)                if(dp[t][j] > dp[t][j-i] + dp[v][i] + i * edge[p].w)                {                    dp[t][j] = dp[t][j-i] + dp[v][i] + i * edge[p].w;                }            }        }    }}int main(){    while(scanf("%d%d%d", &n, &s, &k) != EOF)    {        cnt = 0;        memset(head, -1, sizeof(head));        for(int i = 1; i < n; i++)        {            int a, b, c;            scanf("%d%d%d", &a, &b, &c);            addedge(a, b, c);            addedge(b, a, c);        }        memset(dp, -1, sizeof(dp));        dfs(s, -1);        printf("%d\n", dp[s][k]);    }    return 0;}