[九省联考2018]秘密袭击coat 伪·题解

爆算碾标程实例

不太会多项式……不太会线段树合并

那就只能O(n^2*w^2)爆算+乱搞优化（见代码）

（这里网上都说是O(n*w^2)，我不太明白，也许是我算的不对，望有识之士教我）

愣是卡进luogu最优解第3页

自以为要卡常数，结果卡了好久以后发现是死循环……

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn=2005;
const int modd=64123;

unsigned short int n, k, w;
unsigned short int dp[maxn][maxn], p[maxn];
unsigned short int can[maxn], f[maxn];

struct gra {
    unsigned short int tm, st[maxn*2], nx[maxn*2], to[maxn*2], tt[maxn*2];
    inline void adde(unsigned short int a, unsigned short int b) {
        tm++;
        nx[tm]=st[a];
        st[a]=tm;
        to[tm]=b;
    }
    void dfs(unsigned short int x, unsigned short int fa) {
        //cout<<x<<‘ ‘<<fa<<endl;
        if(fa == f[x]) return;
        else f[x]=fa;
        unsigned short int i, y, j, q, ttt;
        memset(dp[x], 0, sizeof(dp[x]));
        dp[x][p[x]]=1; can[x]=p[x];
        for(i=st[x]; i != 0; i=nx[i]) {
            y=to[i];
            if(y == fa) continue;
            dfs(y, x);
            ttt=can[x];
            for(j=0; j <= can[y]; ++j) {
                for(q=0; q <= ttt; ++q) {
                    if(q+j > k) break;
                    tt[q+j]=((int)tt[q+j]+(int)dp[x][q]*(int)dp[y][j]%modd)%modd;
                    can[x]<q+j?can[x]=q+j:can[x];
                }
            }
            for(j=0; j <= can[x]; ++j) {
                dp[x][j]=((int)dp[x][j]+(int)tt[j])%modd;
                tt[j]=0;
            }
        }
        /*
        cout<<x<<endl;
        for(i=0; i <= k; i++) {
            cout<<dp[x][i]<<‘ ‘;
        }cout<<endl;
        */
        return;
    }
} G;

struct dd {
    unsigned short int num;
    unsigned short int pl;
}d[maxn];
bool cmp(dd a, dd b) {
    return a.num > b.num;
}

int main() {
    //freopen("data5.in", "r", stdin);
    unsigned short int i, ta, j, tb, ans;
    scanf("%hu%hu%hu", &n, &k, &w);
    for(i=1; i <= n; ++i) {
        scanf("%hu", &d[i].num);
        d[i].pl=i;
    }
    sort(d+1, d+1+n, cmp);
    for(i=1; i <= n-1; ++i) {
        scanf("%hu%hu", &ta, &tb);
        G.adde(ta, tb);
        G.adde(tb, ta);
    }
    ans=0;
    for(i=1; i <= n; ++i) {
        p[d[i].pl]=1;
        //cout<<i<<endl;
        if(i < k) continue;
        G.dfs(d[i].pl, d[i].pl);
        ans=((int)ans+(int)d[i].num*(int)dp[d[i].pl][k]%modd)%modd;
        //cout<<i<<endl;
    }
    printf("%hu\n", ans);
    return 0;
}

原文地址：https://www.cnblogs.com/crraphael/p/11488402.html