[HDU 5956] The Elder （斜率DP + 可持久化单调队列）

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HDU - 5956

树上的一个斜率DP裸题

用斜率DP加可持久化单调队列强行搞。
可持久化单调队列就是弹的时候用二分在log时间内确定弹到什么位置
然后加入的时候一次只修改一个，只要备份一下原来的一个值即可
这题注意开一下longlong

#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <iostream>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <cctype>#include <map>#include <set>#include <queue>#include <bitset>#include <string>#include <complex>using namespace std;typedef pair<int,int> Pii;typedef long long LL;typedef unsigned long long ULL;typedef double DBL;typedef long double LDBL;#define MST(a,b) memset(a,b,sizeof(a))#define CLR(a) MST(a,0)#define SQR(a) ((a)*(a))#define PCUT puts("\n----------")const int maxn=1e5+10, maxe=2*maxn;struct Graph{    int ndn, edn, last[maxn];    int u[maxe], v[maxe], w[maxe], nxt[maxe];    void init(int _n){ndn=_n; edn=0; MST(last,-1);}    void adde(int _u, int _v, int _w)    {        u[edn]=_u; v[edn]=_v; w[edn]=_w;        nxt[edn]=last[_u];        last[_u] = edn++;    }};Graph G;int N,P,que[maxn];LL dp[maxn], S[maxn];LL Up(int,int);LL Down(int,int);void dfs(int,int,int,int);int main(){    #ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);    #endif    int T;    scanf("%d", &T);    for(int ck=1; ck<=T; ck++)    {        scanf("%d%d", &N, &P);        G.init(N);        for(int i=1,u,v,w; i<N; i++)        {            scanf("%d%d%d", &u, &v, &w);            G.adde(u,v,w); G.adde(v,u,w);        }        S[1] = 0; dp[1] = -P;        que[0] = 1;        for(int e=G.last[1]; ~e; e=G.nxt[e])        {            S[G.v[e]] = G.w[e];            dfs(G.v[e],1,0,1);        }        LL ans = 0;        for(int i=2; i<=N; i++) ans = max(ans, dp[i]);        printf("%lld\n", ans);    }    return 0;}void dfs(int u, int f, int head, int tail){    int l=head, r=tail-2;    while(l<r)    {        int mid = (l+r)>>1;        if(Up(que[mid+1], que[mid]) <= S[u]*Down(que[mid+1], que[mid])) l=mid+1;        else r=mid;    }    if(l+1<tail && Up(que[l+1], que[l]) <= S[u]*Down(que[l+1], que[l])) l++;    head = l;    dp[u] = dp[que[head]] + SQR(S[u]-S[que[head]]) + P;    LL old, oldp;    l = head+2, r = tail;    while(l<r)    {        int mid = (l+r)>>1;        if(Up(que[mid-1], que[mid-2])*Down(u, que[mid-1]) >= Up(u, que[mid-1])*Down(que[mid-1], que[mid-2])) r = mid;        else l=mid+1;    }    if(head+1<l && Up(que[l-1], que[l-2])*Down(u, que[l-1]) >= Up(u, que[l-1])*Down(que[l-1], que[l-2])) l--;    tail = l;    old = que[tail], oldp = tail;    que[tail++] = u;    for(int e=G.last[u],v; ~e; e=G.nxt[e]) if((v=G.v[e]) != f)    {        S[v] = S[u] + G.w[e];        dfs(v, u, head, tail);    }    que[oldp] = old;}LL Up(int j, int k){return (dp[j] + SQR(S[j]) ) - (dp[k] + SQR(S[k]));}LL Down(int j, int k){return 2*(S[j] - S[k]);}

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