Baltic 2010 Candies

http://www.cppblog.com/jsn1993/archive/2010/05/21/116026.html 上有一个很不错的方法，但可惜最坏情况下区间数还是O(n)级别的（b[1], b[2], ..., b[i]分别为2^i，后面的都取7000, 则所有偶数都可以组合出来，奇数都不可组合出来），因此其动规还是O(N^3)级别的。如果分组优化，把每组的大小设为(sqrt(n*4+1)+1)/2，可以把最坏情况的复杂度降至O(n^2.5)。

 

后来看了官方题解，发现其方法非常巧妙，核心为以下函数：

Algorithm 1 generate((p1 , p2 , p3 , ..., pm ), S)
 if m = 1 then
    S is one of the desired structures
 else
    S1 = add(S, (p1 , p2 , ..., p m/2 ))
    S2 = add(S, (p m/2 +1 , ..., pm ))
    generate((p m/2+1 , ..., pm ), S1 )
    generate((p1 , p2 , ..., p m/2 ), S2 )
 end if

 

参数中S为已经获得的可组合出的数，p1..pm为要加入的包。该算法可在O(B * n * m log m)内运行完毕，并能获得S加上p的所有m-1个元素的子集所能得到的组合结果。

由于官方数据比较厚道，区间数很少，我想到在以上算法的基础上加上区间处理的优化，于是程序运行的效果非常好……

 

/* * $File: p1961.cpp * $Date: Wed Jun 30 09:30:02 2010 +0800 * * Source: Baltic 2010 - candies * */#define INPUT"p1961.in"#define OUTPUT"p1961.out"#include <cstdio>#include <cstring>namespace Solve{const int NPACKAGE_MAX = 105, CAP_MAX = 7000, NINTER_MAX = NPACKAGE_MAX * CAP_MAX, NSTACK_MAX = 10;struct Interval{int left, right;};struct Order{Interval inter[NINTER_MAX];int ninter;inline Order& operator = (const Order &n){ninter = n.ninter;memcpy(inter, n.inter, sizeof(Interval) * ninter);return *this;}};Order stack[NSTACK_MAX];int package[NPACKAGE_MAX], npackage;int remove_pos, remove_val;void add(Order &order, int pkg);void generate(int pkg_left, int pkg_right, int top);int work();void solve(FILE *fin, FILE *fout);}void Solve::solve(FILE *fin, FILE *fout){fscanf(fin, "%d", &npackage);for (int i = 0; i < npackage; i ++)fscanf(fin, "%d", &package[i]);int ans = work();fprintf(fout, "%d %d/n", package[remove_pos], ans);}int Solve::work(){stack[0].ninter = 1;stack[0].inter[0].left = stack[0].inter[0].right = 0;generate(0, npackage - 1, 0);static Order ans;ans.ninter = 1;ans.inter[0].left = ans.inter[0].right = 0;for (int i = 0; i < npackage; i ++)if (i != remove_pos){add(ans, package[i]);add(ans, -package[i]);}for (int i = 0; i < ans.ninter; i ++)if (ans.inter[i].right > 0){if (ans.inter[i].left > 1)return ans.inter[i].left - 1;return ans.inter[i].right + 1;}return -1;}void Solve::generate(int pkg_left, int pkg_right, int top){if (pkg_left == pkg_right){int tot = 0;const Interval *x = stack[top].inter;for (int i = 0; i < stack[top].ninter; i ++)tot += x[i].right - x[i].left + 1;if (tot > remove_val || (tot == remove_val &&package[pkg_left] < package[remove_pos])){remove_val = tot;remove_pos = pkg_left;}return;}top ++;int mid = (pkg_left + pkg_right) >> 1;stack[top] = stack[top - 1];for (int i = pkg_left; i <= mid; i ++)add(stack[top], package[i]);generate(mid + 1, pkg_right, top);stack[top] = stack[top - 1];for (int i = mid + 1; i <= pkg_right; i ++)add(stack[top], package[i]);generate(pkg_left, mid, top);}void Solve::add(Order &order, int pkg){static Interval inter1[NINTER_MAX], inter2[NINTER_MAX];int ninter = order.ninter;memcpy(inter1, order.inter, sizeof(Interval) * ninter);for (int i = 0; i < ninter; i ++){inter2[i].left = inter1[i].left + pkg;inter2[i].right = inter1[i].right + pkg;}int ptr = 0, i = 0, j = 0;Interval cur;if (pkg >= 0)cur = inter1[i ++];else cur = inter2[j ++];while (i < ninter || j < ninter){bool cb = false;if (i < ninter && inter1[i].left <= cur.right + 1){cb = true;if (inter1[i].right > cur.right)cur.right = inter1[i].right;i ++;}if (j < ninter && inter2[j].left <= cur.right + 1){cb = true;if (inter2[j].right > cur.right)cur.right = inter2[j].right;j ++;}if (!cb){order.inter[ptr ++] = cur;if ((i < ninter && inter1[i].left < inter2[j].left) ||j == ninter)cur = inter1[i ++];else cur = inter2[j ++];}}order.inter[ptr ++] = cur;order.ninter = ptr;}int main(){#ifdef STDIOSolve::solve(stdin, stdout);#elseFILE *fin = fopen(INPUT, "r"), *fout = fopen(OUTPUT, "w");Solve::solve(fin, fout);fclose(fin);fclose(fout);#endif}