[SHOI 2008]Debt 循环的债务

Description

题库链接

A 欠 B \(x_1\) 元， B 欠 C \(x_2\) 元， C 欠 A \(x_3\) 元。现每人手上各有若干张 100,50,20,10,5,1 钞票。问至少交换几张钞票才能互相还清债务。

\(1\leq |x_1|,|x_2|,|x_3|\leq 1000\) ，总钱数 \(\leq 1000\)

Solution

考虑 \(DP\) 。

记 \(f_{i,j,k}\) 为当前考虑第 \(i\) 张钞票转移， A 有 \(j\) 元， B 有 \(k\) 元， C 有 \(sum-j-k\) 元。其中 \(sum\) 为三人拥有钞票价值总和。

第 \(i\) 张转移的时候就考虑分钱后每人各有多少张 \(i\) 种钞票。记得去掉不合法的情况。先从钞票面值大到小转移状态会少些。

Code

//It is made by Awson on 2018.2.27
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 1000, w[7] = {0, 100, 50, 20, 10, 5, 1};
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

int x1, x2, x3, sum, INF;
int a[5][7], cnt[7], tol[5], f[8][N+5][N+5];

void work() {
    read(x1), read(x2), read(x3);
    for (int i = 1; i <= 3; i++) for (int j = 1; j <= 6; j++) read(a[i][j]), cnt[j] += a[i][j], tol[i] += a[i][j]*w[j];
    sum += tol[1]+tol[2]+tol[3];
    memset(f, 127/3, sizeof(f)); INF = f[0][0][0]; f[1][tol[1]][tol[2]] = 0;
    for (int i = 1; i <= 6; i++)
    for (int j = 0; j <= sum; j++)
        for (int k = 0; k+j <= sum; k++) {
        if (f[i][j][k] == INF) continue;
        for (int p = 0; p <= cnt[i]; p++)
            for (int q = 0; p+q <= cnt[i]; q++) {
            int p1 = j+(p-a[1][i])*w[i], p2 = k+(q-a[2][i])*w[i], p3 = sum-tol[1]-tol[2]+(cnt[i]-p-q-a[3][i])*w[i];
            if (p1 < 0 || p2 < 0 || p3 < 0) continue;
            if (f[i+1][p1][p2] > f[i][j][k]+(Abs(p-a[1][i])+Abs(q-a[2][i])+Abs(cnt[i]-p-q-a[3][i]))/2)
                f[i+1][p1][p2] = f[i][j][k]+(Abs(p-a[1][i])+Abs(q-a[2][i])+Abs(cnt[i]-p-q-a[3][i]))/2;
            }
        }
    tol[1] -= x1, tol[2] += x1;
    tol[2] -= x2, tol[3] += x2;
    tol[3] -= x3, tol[1] += x3;
    if (tol[1] < 0 || tol[2] < 0 || tol[3] <0 || f[7][tol[1]][tol[2]] == INF) puts("impossible");
    else writeln(f[7][tol[1]][tol[2]]);
}
int main() {
    work(); return 0;
}