# BZOJ1021 SHOI2008循环的债务

dp模拟即可。

d[i][j][k]表示使用前i种面值，1号手里钱为j，2号手里钱为k时最少操作数

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int n = 6;
const int M = 1005;
const int val[n] = {1, 5, 10, 20, 50, 100};
int x1, x2, x3;
int now, tot;
int suma, sumb, dis;
int sum[3], Cnt[n];
int d[2][M][M], cnt[3][n];
int gcd(int a,int b)
{
return b==0?a:gcd(b,a%b);
}
inline void update(int &x, int y){
if (x == -1) x = y;
else x = min(x, y);
}

inline int calc(int i,int a,int b){
return (abs(a - cnt[0][i]) + abs(b - cnt[1][i]) + abs(Cnt[i] - a - b - cnt[2][i])) / 2;
}

int main(){
scanf("%d%d%d", &x1, &x2, &x3);
for (int i = 0; i < 3; ++i){
sum[i] = 0;
for (int j = n - 1; j >= 0; --j){
scanf("%d", cnt[i] + j);
Cnt[j] += cnt[i][j];
sum[i] += cnt[i][j] * val[j];
}
tot += sum[i];
}
int ea = sum[0], eb = sum[1], ec = sum[2];
ea += x3 - x1, eb += x1 - x2, ec += x2 - x3;
if (ea < 0 || eb < 0 || ec < 0 || ea + eb + ec != tot)
{printf("impossible\n");return 0;}
memset(d[0], -1, sizeof(d[0]));
d[0][sum[0]][sum[1]] = 0;
for (int i = 0; i < n; ++i)
{
now = i & 1;
int g=val[i];
for(int j=i+1;j<n;++j)
g=gcd(g,val[j]);
memset(d[now ^ 1], -1, sizeof(d[now ^ 1]));
for (int j = 0; j <= tot; ++j)
{
for (int k = 0; j + k <= tot; ++k)
{
if (d[now][j][k] >= 0)
{
update(d[now ^ 1][j][k], d[now][j][k]);
for (int a = 0; a <= Cnt[i]; ++a)
{
for (int b = 0; a + b <= Cnt[i]; ++b)
{
suma = j + val[i] * (a - cnt[0][i]);
sumb = k + val[i] * (b - cnt[1][i]);
if (suma >= 0 && sumb >= 0 && suma + sumb <= tot)
{
dis = calc(i,a,b);
update(d[now ^ 1][suma][sumb], d[now][j][k] + dis);
}
}
}
}
}
}
}
if(d[n&1][ea][eb]<0) puts("impossible");
else printf("%d\n", d[n & 1][ea][eb]);
return 0;
}

posted @ 2018-01-19 21:34  大奕哥&VANE  阅读(121)  评论(0编辑  收藏  举报