[IOI 1999] 花店橱窗布置

[题目链接]

https://www.luogu.org/problemnew/show/P1854v

[算法]

f[i][j]表示放了前i束花，第i束花放在第j个花瓶中，所能获得的最大美学值

由于要输出方案，我们不妨对于每个状态记一个pre，最后沿着pre数组反推出方案即可

[代码]

#include<bits/stdc++.h>
using namespace std;
#define MAXF 110
const int INF = 2e9;

int i,j,k,now,pos,F,V;
int a[MAXF][MAXF],f[MAXF][MAXF],pre[MAXF][MAXF];
vector< int > path;

int main()
{

        scanf("%d%d",&F,&V);
        for (i = ; i <= F; i++)
        {
                for (j = ; j <= V; j++)
                {
                        scanf("%d",&a[i][j]);
                }
        }
        for (i = ; i <= F; i++)
        {
                for (j = ; j <= V; j++)
                {
                        f[i][j] = -INF;
                }
        }
        for (i = ; i <= F; i++)
        {
                for (j = ; j <= V; j++)
                {
                        for (k = j - ; k >= ; k--)
                        {
                                if (f[i - ][k] + a[i][j] > f[i][j])
                                {
                                        f[i][j] = f[i - ][k] + a[i][j];
                                        pre[i][j] = k;
                                }
                        }
                }
        }
        pos = ;
        for (i = ; i <= V; i++)
        {
                if (f[F][i] > f[F][pos])
                        pos = i;
        }
        printf("%d\n",f[F][pos]);
        now = F;
        while (now)
        {
                path.push_back(pos);
                pos = pre[now][pos];
                now--;
        }
        reverse(path.begin(),path.end());
        for (i = ; i < path.size() - ; i++) printf("%d ",path[i]);
        printf("%d\n",path[path.size() - ]);

        return ;

}