HDU

HDU - 5815

感觉要是值域范围是1000以内， 感觉还是能写出来的。。

考虑dp[ i ][ j ]表示从 1 到 i 路径上的值是 j ， i 这棵子树的最大贡献值。

然后可以发现 j 这维可以离散化， 离散化之后最优值不会变， 然后dp一遍就好了。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0)                     ;

using namespace std;

const int N = 1000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = (int)1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, m;
int hs[N], hscnt;
int ans[N];

LL dp[N][N];
PLI sufMax[N][N];

vector<PII> G[N];
vector<int> V[N];

void dfs(int u, int fa) {
    for(int i = 1; i <= hscnt; i++) {
        dp[u][i] = 0;
    }
    for(auto &e : G[u]) {
        if(e.se == fa) continue;
        dfs(e.se, u);
    }
    for(int i = hscnt, v, pt = 0; i >= 1; i--) {
        while(pt < SZ(V[u]) && V[u][pt] >= hs[i]) pt++;
        dp[u][i] += 1LL * hs[i] * pt;
    }
    for(int i = 1; i <= hscnt; i++) {
        sufMax[u][i] = mk(dp[u][i], i);
    }
    for(int i = hscnt - 1; i >= 1; i--) {
        chkmax(sufMax[u][i], sufMax[u][i + 1]);
    }
    for(int i = 1; i <= hscnt; i++) {
        dp[fa][i] += sufMax[u][i].fi;
    }
}

void getPath(int u, int p, int fa) {
    for(auto &e : G[u]) {
        int id = e.fi, v = e.se;
        if(v == fa) continue;
        ans[id] = hs[sufMax[v][p].se] - hs[p];
        getPath(v, sufMax[v][p].se, u);
    }
}

void init() {
    hscnt = 0;
    for(int i = 1; i <= n; i++) {
        V[i].clear();
        G[i].clear();
    }
}

int main() {
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);

        init();

        for(int i = 1; i < n; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            G[u].push_back(mk(i, v));
            G[v].push_back(mk(i, u));
        }

        hs[++hscnt] = 0;
        for(int i = 1; i <= m; i++) {
            int c, b;
            scanf("%d%d", &c, &b);
            V[c].push_back(b);
            hs[++hscnt] = b;
        }

        sort(hs + 1, hs + 1 + hscnt);
        hscnt = unique(hs + 1, hs + 1 + hscnt) - hs - 1;

        for(int i = 1; i <= n; i++) {
            sort(V[i].rbegin(), V[i].rend());
        }

        dfs(1, 0);

        printf("%lld
", dp[1][1]);

        getPath(1, 1, 0);

        for(int i = 1; i < n; i++) {
            printf("%d%c", ans[i], " 
"[i == n - 1]);
        }
    }
    return 0;
}

/*
*/