제출 #1039867

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1039867		RecursiveCo	봉쇄 시간 (IOI23_closing)	C++17	8 / 100	82 ms	32092 KiB

closing

// CF template, version 3.0

#include <bits/stdc++.h>

using namespace std;

#define improvePerformance ios_base::sync_with_stdio(false); cin.tie(0)
#define getTest int t; cin >> t
#define eachTest for (int _var=0;_var<t;_var++)
#define get(name) int (name); cin >> (name)
#define out(o) cout << (o)
#define getList(cnt, name) vector<int> (name); for (int _=0;_<(cnt);_++) { get(a); (name).push_back(a); }
#define sortl(name) sort((name).begin(), (name).end())
#define rev(name) reverse((name).begin(), (name).end())
#define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
#define decision(b) if (b){out("YES");}else{out("NO");}

//#define int long long int

template <typename T, typename I>
struct segtree {
    int n;
    vector<T> tree;
    vector<I> initial;
    T id;

    segtree(int i_n, vector<I> i_initial, T i_id): n(i_n), initial(i_initial), id(i_id) {
        tree.resize(4 * n);
    }

    T conquer(T left, T right) {
        // write your conquer function
    }

    T value(I inp) {
        // write your value function
    }

    void build(int v, int l, int r) {
        if (l == r) tree[v] = value(initial[l]);
        else {
            int middle = (l + r) / 2;
            build(2 * v, l, middle);
            build(2 * v + 1, middle + 1, r);
            tree[v] = conquer(tree[2 * v], tree[2 * v + 1]);
        }
    }

    void upd(int v, int l, int r, int i, I x) {
        if (l == r) tree[v] = value(x);
        else {
            int middle = (l + r) / 2;
            if (middle >= i) upd(2 * v, l, middle, i, x);
            else upd(2 * v + 1, middle + 1, r, i, x);
            tree[v] = conquer(tree[2 * v], tree[2 * v + 1]);
        }
    }

    T query(int v, int l, int r, int ql, int qr) {
        if (ql <= l && r <= qr) return tree[v];
        else if (r < ql || qr < l) return id;
        int middle = (l + r) / 2;
        T left = query(2 * v, l, middle, ql, qr);
        T right = query(2 * v + 1, middle + 1, r, ql, qr);
        return conquer(left, right);
    }
};

// vector<int>

vector<vector<pair<int, int>>> adjList;
vector<long long> fromX;
vector<long long> fromY;

void dfs(int v, int p, long long d, int which) {
    if (which) fromY[v] = d;
    else fromX[v] = d;
    for (auto pa: adjList[v]) {
        int con = pa.first;
        long long w = pa.second;
        if (con == p) continue;
        dfs(con, v, d + w, which);
    }
}

int max_score(int N, int X, int Y, long long K, vector<int> U, vector<int> V, vector<int> W) {
    adjList.clear();
    adjList.resize(N);
    fromX.clear();
    fromX.resize(N);
    fromY.clear();
    fromY.resize(N);
    forto(N - 1, i) {
        int a = U[i];
        int b = V[i];
        int w = W[i];
        adjList[a].push_back({b, w});
        adjList[b].push_back({a, w});
    }
    dfs(X, -1, 0ll, 0);
    dfs(Y, -1, 0ll, 1);
    int ans = 0;
    if (fromY[X] > 2 * K) {
        vector<long long> paths;
        forto(N, i) {
            paths.push_back(min(fromX[i], fromY[i]));
        }
        sortl(paths);
        long long sum = 0;
        forto(N, i) {
            if (sum + paths[i] > K) break;
            sum += paths[i];
            ans++;
        }
    } else {
        long long C = fromX[Y]; // = fromY[X]
        // CASE 1: Stuff gets taken twice on both sides.
        // Everything in the middle gets taken twice.
        // Greedy/bruteforce on the sides.
        int ans1 = 0;
        long long sum1 = 0;
        int middle1 = 0;
        vector<long long> sides1;
        forto(N, i) {
            long long val = max(fromX[i], fromY[i]);
            if (val < C) sum1 += val, middle1++;
            else sides1.push_back(min(fromX[i], fromY[i]));
        }
        sortl(sides1);
        int s1 = sides1.size();
        if (sum1 <= K) ans1 = max(ans1, 2 * middle1);
        forto(s1, i) {
            if (sum1 + sides1[i] > K) break;
            sum1 += sides1[i];
            long long extral = (K - sum1) / C;
            int extra;
            if (extral > (long long) i + 1) extra = i + 1;
            else extra = (int) extral;
            ans1 = max(ans1, 2 * middle1 + (i + 1) + extra);
        }

        // CASE 2: Stuff gets taken twice on one of the two sides.
        // Everything in the middle gets taken once.
        // Suppose wlog the twice-taken stuff occurs on Y's side, then some suffix spanning to at least the halfway point of the middle must get taken twice.
        // CASE 2.1: On Y's side.
        int ans21 = 0;
        long long sum21 = 0;
        int middle21 = 0;
        vector<long long> delta21;
        vector<long long> side21;
        forto(N, i) {
            long long val = min(fromX[i], fromY[i]);
            if (max(fromX[i], fromY[i]) < C) sum21 += val, middle21++;
            if (max(fromX[i], fromY[i]) < C) {
                if (2 * fromY[i] <= C) sum21 += max(fromX[i], fromY[i]) - val, middle21++;
                else delta21.push_back(fromY[i] - fromX[i]);
            } else if (fromX[i] <= fromY[i]) {
                delta21.push_back(fromX[i]);
            } else {
                side21.push_back(fromY[i]);
            }
        }
        sortl(delta21);
        sortl(side21);
        if (sum21 <= K) ans21 = max(ans21, middle21);
        int ds21 = delta21.size();
        int s21 = side21.size();
        forto(ds21, i) {
            if (sum21 + delta21[i] > K) break;
            sum21 += delta21[i];
            long long here = sum21;
            forto(s21, j) {
                if (here + side21[i] > K) break;
                here += side21[i];
                long long extral = (K - here) / C;
                int extra;
                if (extral > (long long) j + 1) extra = j + 1;
                else extra = (int) extral;
                ans21 = max(ans21, middle21 + (i + 1) + (j + 1) + extra);
            }
        }

        // CASE 2.2: On X's side.
        int ans22 = 0;
        long long sum22 = 0;
        int middle22 = 0;
        vector<long long> delta22;
        vector<long long> side22;
        forto(N, i) {
            long long val = min(fromX[i], fromY[i]);
            if (max(fromX[i], fromY[i]) < C) sum22 += val, middle22++;
            if (max(fromX[i], fromY[i]) < C) {
                if (2 * fromX[i] <= C) sum22 += max(fromX[i], fromY[i]) - val, middle22++;
                else delta22.push_back(fromX[i] - fromY[i]);
            } else if (fromY[i] <= fromX[i]) {
                delta22.push_back(fromY[i]);
            } else {
                side22.push_back(fromX[i]);
            }
        }
        sortl(delta22);
        sortl(side22);
        if (sum22 <= K) ans22 = max(ans22, middle22);
        int ds22 = delta22.size();
        int s22 = side22.size();
        forto(ds22, i) {
            if (sum22 + delta22[i] > K) break;
            sum22 += delta22[i];
            long long here = sum22;
            forto(s22, j) {
                if (here + side22[i] > K) break;
                here += side22[i];
                long long extral = (K - here) / C;
                int extra;
                if (extral > (long long) j + 1) extra = j + 1;
                else extra = (int) extral;
                ans22 = max(ans22, middle22 + (i + 1) + (j + 1) + extra);
            }
        }

        // CASE 3: Nothing on the sides gets taken twice.
        // Then we greedy on the sides while solving the problem in the middle.
        // Yeah O(n^2log(n)) passes. Pretty sure.
        int ans3 = 0;
        vector<long long> sides3;
        vector<long long> midsmall3;
        vector<long long> middelta3;
        long long midsum3 = 0;
        forto(N, i) {
            long long val = max(fromX[i], fromY[i]);
            if (val < C) {
                midsmall3.push_back(min(fromX[i], fromY[i]));
                middelta3.push_back(val - min(fromX[i], fromY[i]));
                midsum3 += min(fromX[i], fromY[i]);
            } else {
                sides3.push_back(min(fromX[i], fromY[i]));
            }
        }
        sortl(sides3);
        sortl(midsmall3);
        sortl(middelta3);
        int s3 = sides3.size();
        int ms3 = midsmall3.size();
        int md3 = middelta3.size();
        long long sum3 = 0;
        forto(s3, i) {
            if (sum3 + sides3[i] > K) break;
            sum3 += sides3[i];
            long long tot = sum3;
            forto(ms3, j) {
                if (tot + midsmall3[j] > K) break;
                tot += midsmall3[j];
                ans3 = max(ans3, (i + 1) + (j + 1));
            }
            if (sum3 + midsum3 <= K) {
                long long higher = sum3 + midsum3;
                forto(md3, j) {
                    if (higher + middelta3[j] > K) break;
                    higher += middelta3[j];
                    ans3 = max(ans3, (i + 1) + ms3 + (j + 1));
                }
            }
        }

        ans = max({ans1, ans21, ans22, ans3});
    }
    return ans;
}

컴파일 시 표준 에러 (stderr) 메시지

closing.cpp: In function 'int max_score(int, int, int, long long int, std::vector<int>, std::vector<int>, std::vector<int>)':
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:93:5: note: in expansion of macro 'forto'
   93 |     forto(N - 1, i) {
      |     ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:105:9: note: in expansion of macro 'forto'
  105 |         forto(N, i) {
      |         ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:110:9: note: in expansion of macro 'forto'
  110 |         forto(N, i) {
      |         ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:124:9: note: in expansion of macro 'forto'
  124 |         forto(N, i) {
      |         ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:132:9: note: in expansion of macro 'forto'
  132 |         forto(s1, i) {
      |         ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:151:9: note: in expansion of macro 'forto'
  151 |         forto(N, i) {
      |         ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:168:9: note: in expansion of macro 'forto'
  168 |         forto(ds21, i) {
      |         ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'j' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:172:13: note: in expansion of macro 'forto'
  172 |             forto(s21, j) {
      |             ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:189:9: note: in expansion of macro 'forto'
  189 |         forto(N, i) {
      |         ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:206:9: note: in expansion of macro 'forto'
  206 |         forto(ds22, i) {
      |         ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'j' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:210:13: note: in expansion of macro 'forto'
  210 |             forto(s22, j) {
      |             ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:229:9: note: in expansion of macro 'forto'
  229 |         forto(N, i) {
      |         ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:246:9: note: in expansion of macro 'forto'
  246 |         forto(s3, i) {
      |         ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'j' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:250:13: note: in expansion of macro 'forto'
  250 |             forto(ms3, j) {
      |             ^~~~~
closing.cpp:15:35: warning: unnecessary parentheses in declaration of 'j' [-Wparentheses]
   15 | #define forto(name, var) for (int (var) = 0; (var) < (name); (var)++)
      |                                   ^
closing.cpp:257:17: note: in expansion of macro 'forto'
  257 |                 forto(md3, j) {
      |                 ^~~~~

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