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 STDIO Solve::solve(stdin, stdout); #else FILE *fin = fopen(INPUT, "r"), *fout = fopen(OUTPUT, "w"); Solve::solve(fin, fout); fclose(fin); fclose(fout); #endif }