제출 #162416

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
162416		tri	팀들 (IOI15_teams)	C++14	34 / 100	4051 ms	248676 KiB

teams

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;

#define pb push_back
#define f first
#define s second

namespace debug {
    const int DEBUG = true;

    template<class T1, class T2>
    void pr(const pair<T1, T2> &x);

    template<class T, size_t SZ>
    void pr(const array<T, SZ> &x);

    template<class T>
    void pr(const vector<T> &x);

    template<class T>
    void pr(const set<T> &x);

    template<class T1, class T2>
    void pr(const map<T1, T2> &x);

    template<class T>
    void pr(const T &x) { if (DEBUG) cout << x; }

    template<class T, class... Ts>
    void pr(const T &first, const Ts &... rest) { pr(first), pr(rest...); }

    template<class T1, class T2>
    void pr(const pair<T1, T2> &x) { pr("{", x.f, ", ", x.s, "}"); }

    template<class T>
    void prIn(const T &x) {
        pr("{");
        bool fst = 1;
        for (auto &a : x) {
            pr(fst ? "" : ", ", a), fst = 0;
        }
        pr("}");
    }

    template<class T, size_t SZ>
    void pr(const array<T, SZ> &x) { prIn(x); }

    template<class T>
    void pr(const vector<T> &x) { prIn(x); }

    template<class T>
    void pr(const set<T> &x) { prIn(x); }

    template<class T1, class T2>
    void pr(const map<T1, T2> &x) { prIn(x); }

    void ps() { pr("\n"), cout << flush; }

    template<class Arg, class... Args>
    void ps(const Arg &first, const Args &... rest) {
        pr(first, " ");
        ps(rest...);
    }
}
using namespace debug;


bool success = true;
int standardNx;

namespace PSegTr {
    static const int MAXN = 500000, NODES = 40 * MAXN;
    // different definition of lz to nicely implement persistence
    // lz is only for children, node which it is on already contains updated value
    int lC[NODES], rC[NODES], tr[NODES];
    int nx = 0;

    int cpy(int o) {
        int n = nx++;
        lC[n] = lC[o], rC[n] = rC[o], tr[n] = tr[o];
        return n;
    }

    // q does not create any new nodes
    int q(int i, int l, int r, int s, int e) {
        if (e < l || r < s) {
            return 0;
        }
        if (s <= l && r <= e) {
            return tr[i];
        }
        int mid = (l + r) / 2;
        return q(lC[i], l, mid, s, e) +
               q(rC[i], mid + 1, r, s, e);
    }


    // returns index of updated node
    int u(int i, int l, int r, int x, int d) {
        i = cpy(i);
        if (l == r) {
            tr[i] += d;
            return i;
        }
        int mid = (l + r) / 2;
        if (x <= mid) {
            lC[i] = u(lC[i], l, mid, x, d);
        } else {
            rC[i] = u(rC[i], mid + 1, r, x, d);
        }
        tr[i] = tr[lC[i]] + tr[rC[i]];
        return i;
    }

    int bEmpty(int l, int r) {
        int i = nx++;
        if (l == r) {
            tr[i] = 0;
            return i;
        }
        int mid = (l + r) / 2;
        lC[i] = bEmpty(l, mid);
        rC[i] = bEmpty(mid + 1, r);
        tr[i] = tr[lC[i]] + tr[rC[i]];
        assert(tr[i] == 0);
        return i;
    }

    int clr() {
        memset(lC, 0, sizeof(lC)), memset(rC, 0, sizeof(rC)), memset(tr, 0, sizeof(tr));
        nx = 0;
    }

    int useK(int n, int u, int l, int r, int k) {
//        ps("l, r, k:", l, r, k);
        if(k == 0){
            return u;
        }

        int rem = tr[n] - tr[u];
        if (rem < k) {
            success = false;
            return u;
        }

        u = cpy(u);
        tr[u] += k;
        if (l == r) {
            return u;
        }

        int remL = tr[lC[n]] - tr[lC[u]];

        int mid = (l + r) / 2;
        if (remL >= k) {
            lC[u] = useK(lC[n], lC[u], l, mid, k);
        } else {
            lC[u] = lC[n]; // all of it is used, interesting how you can copy subtrees with persistence
            rC[u] = useK(rC[n], rC[u], mid + 1, r, k - max(0, remL));
        }
        return u;
    }

    // cut is exclusive
    int remB(int i, int l, int r, int cut) {
        i = cpy(i);
        if (l == r) {
            tr[i] = 0;
            return i;
        }
        int mid = (l + r) / 2;
        if (cut <= mid) {
            lC[i] = remB(lC[i], l, mid, cut);
        } else {
            lC[i] = cpy(lC[i]);
            lC[i] = 0; // don't need to worry about children as we shouldn't ever descend into it again
            rC[i] = remB(rC[i], mid + 1, r, cut);
        }
        tr[i] = tr[lC[i]] + tr[rC[i]];
        return i;
    }

    vi trees;

    void uTrees(int x, int d) {
        trees.pb(u(trees.back(), 1, MAXN, x, d));
    }

    int qTrees(int t, int s, int e) {
        return q(trees[t], 1, MAXN, s, e);
    }

    vector<vi> layers;

    int printTr(int x, int l, int r, int d) {
        layers[d].pb(tr[x]);
        if (l != r) {
            int mid = (l + r) / 2;
            printTr(lC[x], l, mid, d + 1);
            printTr(rC[x], mid + 1, r, d + 1);
        }
    }

    void printTr(int x) {
        layers.clear();
        layers.resize(10);
        printTr(x, 1, MAXN, 0);
        for (int i = 0; i < 10; i++) {
            ps(layers[i]);
        }
    }
}

using namespace PSegTr;

int N;
vector<pi> pts;

int init(int iN, int a[], int b[]) {
    N = iN;
    for (int i = 0; i < N; i++) {
        pts.pb({a[i], b[i]});
    }
    pts.pb({-1, -1});
    sort(pts.begin(), pts.end());

    clr();
    trees.pb(bEmpty(1, MAXN));

    for (int i = 1; i <= N; i++) {
        uTrees(pts[i].s, +1);
    }

//    printTr(trees[1]);

    standardNx = nx;
}

int can(int M, int k[]) {
    success = true;
    nx = standardNx; // save space by reusing elements
    sort(k, k + M);
//    ps(allK);

    int used = bEmpty(1, MAXN);

    for (int i = 0; i < M; i++) {
        int rMostValid = upper_bound(pts.begin(), pts.end(), make_pair(k[i], (int) 1e8)) - pts.begin() - 1;
        int options = trees[rMostValid];

//        ps(rMostValid);
//        ps("before cut");
//        printTr(options);

        options = k[i] - 1 > 0 ? remB(options, 1, MAXN, k[i] - 1) : options;
//        ps("options", i);
//        printTr(options);

        used = k[i] - 1 > 0 ? remB(used, 1, MAXN, k[i] - 1) : used;

//        ps("k[i]:", k[i]);
        used = useK(options, used, 1, MAXN, k[i]);

        if (success == false) {
//            ps("failed on:", i);
            return 0;
        }

//        ps("used");
//        printTr(used);
    }
    return 1;
}

컴파일 시 표준 에러 (stderr) 메시지

teams.cpp: In function 'int PSegTr::clr()':
teams.cpp:142:5: warning: no return statement in function returning non-void [-Wreturn-type]
     }
     ^
teams.cpp: In function 'int PSegTr::printTr(int, int, int, int)':
teams.cpp:212:5: warning: no return statement in function returning non-void [-Wreturn-type]
     }
     ^
teams.cpp: In function 'int init(int, int*, int*)':
teams.cpp:247:1: warning: no return statement in function returning non-void [-Wreturn-type]
 }
 ^
teams.cpp: In function 'int can(int, int*)':
teams.cpp:258:104: warning: conversion to 'int' from '__gnu_cxx::__normal_iterator<std::pair<int, int>*, std::vector<std::pair<int, int> > >::difference_type {aka long int}' may alter its value [-Wconversion]
         int rMostValid = upper_bound(pts.begin(), pts.end(), make_pair(k[i], (int) 1e8)) - pts.begin() - 1;
                          ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~

• Subtask 1: 21 / 21
• Subtask 2: 13 / 13
• Subtask 3: 0 / 43
• Subtask 4: 0 / 23