bzoj4033 [HAOI2015]树上染色

Description

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有一棵点数为N的树，树边有边权。给你一个在0~N之内的正整数K，你要在这棵树中选择K个点，将其染成黑色，并
将其他的N-K个点染成白色。将所有点染色后，你会获得黑点两两之间的距离加上白点两两之间距离的和的收益。
问收益最大值是多少。

输入保证所有点之间是联通的。
N<=2000,0<=K<=N

来自 https://www.lydsy.com/JudgeOnline/problem.php?id=4033

Solution

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卧槽原来我做过这样的题目？！！
直接做不好搞，可以考虑按边统计贡献。设f[I,j]表示以i为根的子树中分配j个黑点所有边的贡献和，枚举一个儿子k再枚举分配的黑点，一条边的贡献yy一下不就出来了
非常naive的代码风格，大概是去年八月份写的

Code

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#include 
#include 
#include 
#include 
#include 
#include 
#define rep(i, st, ed) for (int i = st; i <= ed; i += 1)
#define drp(i, st, ed) for (int i = st; i >= ed; i -= 1)
#define erg(i, st) for (int i = ls[st]; i; i = e[i].next)
#define fill(x, t) memset(x, t, sizeof(x))
#define max(x, y) (x)>(y)?(x):(y)
#define min(x, y) (x)<(y)?(x):(y)
#define ll long long
#define db double
#define INF 0x3f3f3f3f
#define N 2001
#define E 5001
#define L 2001
struct edge{int x, y, w, next;}e[E];
ll f[N][L], size[N], fa[N], ls[N];
int edgeCnt = 0, ans = 0;
inline ll read(){
    ll x = 0; char ch = getchar();
    while (ch < '0' || ch > '9'){ch=getchar();}
    while (ch <= '9' && ch >= '0'){x = x * 10 + ch - '0'; ch = getchar();}
    return x;
}
inline void addEdge(int x, int y, int w){
    e[++ edgeCnt] = (edge){x, y, w, ls[x]}; ls[x] = edgeCnt;
    e[++ edgeCnt] = (edge){y, x, w, ls[y]}; ls[y] = edgeCnt;
}
inline void dfs1(int now, int m){
    rep(i, 1, m){f[now][i]=-1;}
    size[now] = 1;
    erg(i, now){
        if (e[i].y != fa[now]){
            fa[e[i].y] = now;
            dfs1(e[i].y, m);
            size[now] += size[e[i].y];
        }
    }
}
inline void dfs2(int now, int m){
    erg(i, now){
        if (e[i].y != fa[now]){
            dfs2(e[i].y, m);
        }
    }
    f[now][0]=f[now][1]=0;
    erg(k, now){
        int y = e[k].y;
        if (fa[y] == now){
            int lzh = min(m, size[now]);
            drp(i, lzh, 0){
                int lim = min(i,size[y]);
                rep(j, 0, lim){
                    if (f[now][i-j] != -1 && f[y][j] != -1){
                        f[now][i]=max(f[now][i],f[now][i-j]+f[y][j]+e[k].w*(j*(m-j)+(size[y]-j)*(size[1]-size[y]-m+j)));
                    }
                }
            }
        }
    }
}
int main(void){
/*  freopen("color.in","r",stdin);
    freopen("color.out","w",stdout);
*/  int n = read();
    int m = read();
    rep(i, 2, n){
        int x = read();
        int y = read();
        int w = read();
        addEdge(x, y ,w);
    }
    fill(f, -1);
    dfs1(1, m);
    dfs2(1, m);
    printf("%lld\n", f[1][m]);
    return 0;
}