hdu4003

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很有有意思的树形DP

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题意：

给一棵n个结点的树, 结点编号为1~n,

每条边都有长度. 有k个机器人从S点出发,

问让机器人遍历所有边,最少路程和是多少?

1<= n <=10000, 1 <= S <= n, 1 <= k <= 10

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如果让一个机器人从根出发,遍历所有树之后再回到根,
那么花费一定是所有边权之和的两倍,因为每条边都走了两次。

而这道题中,机器人遍历完之后,并不需要走回出发点, 所以有些边只走了一次就可以了。
考虑一种简单情形，如果用1台机器人走,权值和最大的根到叶子路径只走一次是最优的。
最小花费 = sum*2 – {根到叶子路径的最大权值和}

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可以有这样一个思路：
我们可以让k台机器人减少的花费尽量大.

对于i节点,我们可以选择派1~k个机器人去走
需要注意,如果派x个机器人走向某个子节点v,
那么E(i,v)就会被走了x次, 花费了x*w(i, v).
而原始的sum中每条边只走了两次,
所以走E(i,v)的花费减少了 2*w(i,v) - x*w(i,v)

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状态表示: f(i,j) 表示子树i用j个机器人最多可以减少的花费

状态转移式:
f(i,j)= max{f(i,j-x)+f(v,x)+2*w(i,x)-k*w(i,x)}
(1<=x<=j 且 v是i的儿子节点)

最终答案 ans = sum * 2 - f(S, k)

时间复杂度：O(n*k^2)

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dp(i,j)表示用j个机器人遍历i的子树所花费的最小费用

我们可以将机器人数量视为背包容积。
将i结点的每棵子树视为一组物品，并且每组物品中至少要选一个物品。

这样做背包就可以啦！~

状态转移方程为:
Cost[0]=dp(i,j)+dp(son,0)+2∗w(i,son)
Cost[1]=min{dp(son,x)+dp(i,j−x)+k∗w(i,son)}(1<=x<=j)
dp(i,j)=min{Cost[0],Cost[1]}

当然实现时可以写得更简单。

时间复杂度：O(n*k^2)

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#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <string>#include <map>#include <vector>#include <stack>#include <queue>#include <utility>#include <iostream>#include <algorithm>const int maxn = 10005, maxk = 12, INF = 0x3f3f3f3f;template<class Num>void read(Num &x){    char c; int flag = 1;    while((c = getchar()) < '0' || c > '9')        if(c == '-') flag *= -1;    x = c - '0';    while((c = getchar()) >= '0' && c <= '9')        x = (x<<3) + (x<<1) + (c-'0');    x *= flag;    return;}template<class Num>void write(Num x){    if(x < 0) putchar('-'), x = -x;    static char s[20];int sl = 0;    while(x) s[sl++] = x%10 + '0',x /= 10;    if(!sl) {putchar('0');return;}    while(sl) putchar(s[--sl]);}struct Edge{    int v, w, next;    Edge(int v = 0,int w = 0,int next = 0):v(v),w(w),next(next){}}edge[maxn<<1];int el, head[maxn];int n, S, R;int ans, sum;int dp[maxn][maxk];void NewEdge(int u,int v,int w){    edge[++el] = Edge(v, w, head[u]), head[u] = el;}void clear(){    ans = el = sum = 0;    memset(head, 0, sizeof(head));    memset(dp, 0, sizeof(dp));}void dfs(int a,int fa){    for(int i = head[a], p, w; i ; i = edge[i].next)        if((p = edge[i].v) != fa)         {            dfs(p, a), w = edge[i].w;            for(int j = R; j > 0; j--)                for(int k = 1; k <= j; k++)                {                    int tmp = dp[a][j-k] + dp[p][k] + (2 - k)*w;                    dp[a][j] = std::max(dp[a][j], tmp);                }        }}int main(){#ifndef ONLINE_JUDGE    freopen("hdu4003.in","r",stdin);    freopen("hdu4003.out","w",stdout);#endif    while(std::cin >> n)    {        clear(), std::cin >> S >> R;        for(int i = 1, u, v, w; i < n; i++)        {            read(u), read(v), read(w), sum += w;            NewEdge(u, v, w), NewEdge(v, u, w);        }        dfs(S, 0);        for(int i = 1; i <= R; i++)            ans = std::max(dp[S][i], ans);        ans = (sum<<1) - ans;        std::cout << ans << std::endl;    }#ifndef ONLINE_JUDGE    fclose(stdin);    fclose(stdout);#endif    return 0;}

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#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <string>#include <map>#include <vector>#include <stack>#include <queue>#include <utility>#include <iostream>#include <algorithm>const int maxn = 10005, maxk = 12, INF = 0x3f3f3f3f;template<class Num>void read(Num &x){    char c; int flag = 1;    while((c = getchar()) < '0' || c > '9')        if(c == '-') flag *= -1;    x = c - '0';    while((c = getchar()) >= '0' && c <= '9')        x = (x<<3) + (x<<1) + (c-'0');    x *= flag;    return;}template<class Num>void write(Num x){    if(x < 0) putchar('-'), x = -x;    static char s[20];int sl = 0;    while(x) s[sl++] = x%10 + '0',x /= 10;    if(!sl) {putchar('0');return;}    while(sl) putchar(s[--sl]);}struct Edge{    int v, w, next;    Edge(int v = 0,int w = 0,int next = 0):v(v),w(w),next(next){}}edge[maxn<<1];int el, head[maxn];int n, S, R;int ans, sum;int dp[maxn][maxk];void NewEdge(int u,int v,int w){    edge[++el] = Edge(v, w, head[u]), head[u] = el;}void clear(){    ans = el = sum = 0;    memset(head, 0, sizeof(head));    memset(dp, 0, sizeof(dp));}void dfs(int a,int fa){    for(int i = head[a], p, w; i ; i = edge[i].next)        if((p = edge[i].v) != fa)         {            dfs(p, a), w = edge[i].w;            for(int j = R; j >= 0; j--)            {                dp[a][j] = dp[a][j] + dp[p][0] + (w<<1);                    for(int k = 1; k <= j; k++)                   dp[a][j] = std::min(dp[a][j], dp[a][j-k] + dp[p][k] + k*w);            }//          dp[a][0] += dp[p][0] + (w<<1);        }}int main(){#ifndef ONLINE_JUDGE    freopen("hdu4003.in","r",stdin);    freopen("hdu4003.out","w",stdout);#endif    while(std::cin >> n)    {        clear(), std::cin >> S >> R;        for(int i = 1, u, v, w; i < n; i++)        {            read(u), read(v), read(w);            NewEdge(u, v, w), NewEdge(v, u, w);        }        dfs(S, 0);        std::cout << dp[S][R] << std::endl;    }#ifndef ONLINE_JUDGE    fclose(stdin);    fclose(stdout);#endif    return 0;}