luogu1064 金明的预算方案

  这道题我就想说一点:审题!附件只有2个!钱是10的整数倍,不是100的整数倍!

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAX_OBJ = 60, MAX_V = 5000, MINF = 0xcfcfcfcf;
int ObjCnt[MAX_OBJ], V[MAX_OBJ][5], W[MAX_OBJ][5];
int F[MAX_V];
int TotV, TotObj;

int Dp()
{
    memset(F, MINF, sizeof(F));
    F[0] = 0;
    for (int i = 1; i <= TotObj; i++)
        for (int j = TotV; j >= 0; j--)
            for (int k = 1; k <= ObjCnt[i]; k++)
                if (V[i][k] <= j)
                    F[j] = max(F[j], F[j - V[i][k]] + W[i][k]);
    int ans = 0;
    for (int j = 1; j <= TotV; j++)
        ans = max(ans, F[j]);
    return ans;
}

int main()
{
    scanf("%d%d", &TotV, &TotObj);
    for (int i = 1; i <= TotObj; i++)
        ObjCnt[i] = 1;
    TotV /= 10;
    for (int i = 1; i <= TotObj; i++)
    {
        int v, w, group;
        scanf("%d%d%d", &v, &w, &group);
        v /= 10;
        if (group == 0)
        {
            V[i][1] = v;
            W[i][1] = w * v;
        }
        else
        {
            ObjCnt[group]++;
            V[group][ObjCnt[group]] = v;
            W[group][ObjCnt[group]] = w * v;
        }
    }
    for (int i = 1; i <= TotObj; i++)
    {
        if (ObjCnt[i] == 1 && !V[i][1])
            ObjCnt[i] = 0;
        else if (ObjCnt[i] == 2)
        {
            V[i][2] += V[i][1];
            W[i][2] += W[i][1];
        }
        else if (ObjCnt[i] == 3)
        {
            ObjCnt[i]++;
            V[i][4] = V[i][1] + V[i][2] + V[i][3];
            W[i][4] = W[i][1] + W[i][2] + W[i][3];
            V[i][2] += V[i][1];
            W[i][2] += W[i][1];
            V[i][3] += V[i][1];
            W[i][3] += W[i][1];
        }
    }
    printf("%d\n", Dp() * 10);
    return 0;
}

  

posted @ 2018-08-06 22:26  headboy2002  阅读(66)  评论(0编辑  收藏  举报