vijos P1426兴奋剂检查 多维费用背包问题的hash

Posted 2020-08-31 stupid_one

https://vijos.org/p/1426

这是个好题，容易想到用dp[i][v1][v2][v3][v4][v5]表示在前i个物品中，各种东西的容量是那个的时候，能产生的最大价值。

时间不会TLE,但是会MLE.所以就需要把那5维状态进行hash

其实就是对这个排列进行一个hash。

newState: v1, v2, v3, v4, v5这个排列。

oldState: v1 - w[1], v2 - w[2], v3 - w[3], v4 - w[4], v5 - w[5]

这个排列。

然后要进行hash。我写了一个hash。TLE 两组数据。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define ios ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 200 + 20;
int dp[5000000 + 20];
int limit[maxn];
int w[maxn][10];
int val[maxn];
int togive;
const int seed = 131;
typedef unsigned long long int ULL;
ULL powseed[10];
int first[100003 + 2];
ULL Edge[5000000 + 20];
int tonext[5000000 + 20];
int id[5000000 + 20];
int num;
int toadd(ULL val) {
    int u = val % 100003;
    for (int i = first[u]; i; i = tonext[i]) {
        if (val == Edge[i]) return id[i];
    }
    ++num;
    tonext[num] = first[u];
    first[u] = num;
    Edge[num] = val;
    id[num] = ++togive;
    return togive;
}
int tohash(int a, int b, int c, int d, int e) {
    ULL val = a * powseed[1] + b * powseed[2] +
              c * powseed[3] + d * powseed[4] + e * powseed[5];
    return toadd(val);
}
//适用于正负整数
template <class T>
inline bool fast_in(T &ret) {
    char c;
    int sgn;
    if(c = getchar(), c == EOF) return 0; //EOF
    while(c != \'-\' && (c < \'0\' || c > \'9\')) c = getchar();
    sgn = (c == \'-\') ? -1 : 1;
    ret = (c == \'-\') ? 0 : (c - \'0\');
    while(c = getchar(), c >= \'0\' && c <= \'9\') ret = ret * 10 + (c - \'0\');
    ret *= sgn;
    return 1;
}
void work() {
    int n, m;
//    cin >> n >> m;
//    scanf("%d%d", &n, &m);
    fast_in(n);
    fast_in(m);
    for (int i = 1; i <= m; ++i) {
//        cin >> limit[i];
//        scanf("%d", &limit[i]);
        fast_in(limit[i]);
    }
    for (int i = 1; i <= n; ++i) {
//        cin >> val[i];
//        scanf("%d", &val[i]);
        fast_in(val[i]);
        for (int j = 1; j <= m; ++j) {
//            cin >> w[i][j];
//            scanf("%d", &w[i][j]);
            fast_in(w[i][j]);
        }
    }
//    cout << tohash(1, 2, 3, 4, 5) << endl;
//    cout << tohash(1, 2, 4, 3, 5) << endl;
//    cout << tohash(1, 2, 3, 4, 5) << endl;
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
        for (int a1 = limit[1]; a1 >= w[i][1]; --a1) {
            for (int a2 = limit[2]; a2 >= w[i][2]; --a2) {
                for (int a3 = limit[3]; a3 >= w[i][3]; --a3) {
                    for (int a4 = limit[4]; a4 >= w[i][4]; --a4) {
                        for (int a5 = limit[5]; a5 >= w[i][5]; --a5) {
                            int now = tohash(a1, a2, a3, a4, a5);
                            int pre = tohash(a1 - w[i][1], a2 - w[i][2], a3 - w[i][3], a4 - w[i][4], a5 - w[i][5]);
                            dp[now] = max(dp[now], dp[pre] + val[i]);
                            ans = max(ans, dp[now]);
                        }
                    }
                }
            }
        }
    }
    printf("%d\\n", ans);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    powseed[0] = 1;
    for (int i = 1; i <= 5; ++i) {
        powseed[i] = powseed[i - 1] * seed;
    }
    work();
    return 0;
}

View Code

 

题解的那个hash我真看不懂。不是看不懂，是不理解。

其实他的意思类似于a * 1000000 + b * 100000 + c * 1000 + e * 100 * d * 10

这样类似的。但是明显这个值太大了。他就乘上了limit[i]，这个我不能证明了，

好像又不会重复，数字又小。ORZ

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 200 + 20;
int dp[5000000 + 20];
int limit[maxn];
int w[maxn][10];
int val[maxn];

int tohash(int a, int b, int c, int d, int e) {
    return (a * (limit[2] + 1) * (limit[3] + 1) * (limit[4] + 1) * (limit[5] + 1)
            + b * (limit[3] + 1) * (limit[4] + 1) * (limit[5] + 1)
            + c * (limit[4] + 1) * (limit[5] + 1) + d * (limit[5] + 1) + e);
}
//适用于正负整数
template <class T>
inline bool fast_in(T &ret) {
    char c;
    int sgn;
    if(c = getchar(), c == EOF) return 0; //EOF
    while(c != \'-\' && (c < \'0\' || c > \'9\')) c = getchar();
    sgn = (c == \'-\') ? -1 : 1;
    ret = (c == \'-\') ? 0 : (c - \'0\');
    while(c = getchar(), c >= \'0\' && c <= \'9\') ret = ret * 10 + (c - \'0\');
    ret *= sgn;
    return 1;
}
void work() {
    int n, m;
//    cin >> n >> m;
//    scanf("%d%d", &n, &m);
    fast_in(n);
    fast_in(m);
    for (int i = 1; i <= m; ++i) {
//        cin >> limit[i];
//        scanf("%d", &limit[i]);
        fast_in(limit[i]);
    }
    for (int i = 1; i <= n; ++i) {
//        cin >> val[i];
//        scanf("%d", &val[i]);
        fast_in(val[i]);
        for (int j = 1; j <= m; ++j) {
//            cin >> w[i][j];
//            scanf("%d", &w[i][j]);
            fast_in(w[i][j]);
        }
    }
//    cout << tohash(1, 2, 3, 4, 5) << endl;
//    cout << tohash(1, 2, 4, 3, 5) << endl;
//    cout << tohash(1, 2, 3, 4, 5) << endl;
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
        for (int a1 = limit[1]; a1 >= w[i][1]; --a1) {
            for (int a2 = limit[2]; a2 >= w[i][2]; --a2) {
                for (int a3 = limit[3]; a3 >= w[i][3]; --a3) {
                    for (int a4 = limit[4]; a4 >= w[i][4]; --a4) {
                        for (int a5 = limit[5]; a5 >= w[i][5]; --a5) {
                            int now = tohash(a1, a2, a3, a4, a5);
                            int pre = tohash(a1 - w[i][1], a2 - w[i][2], a3 - w[i][3], a4 - w[i][4], a5 - w[i][5]);
                            dp[now] = max(dp[now], dp[pre] + val[i]);
                            ans = max(ans, dp[now]);
                        }
                    }
                }
            }
        }
    }
    printf("%d\\n", ans);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}

View Code

 

感觉我的hash才是正确的打开方式