1625 夹克爷发红包 贪心 + 暴力 ＋ 思维

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1625&judgeId=203076

一开始的时候，还想着枚举每一行的总和，和每一列的总和，然后选一个最小的来变就好。

然后wa了，只剩下一组数据，然后看到还是大数据，调试不了。

然后上网搜，然后跪了。

我那种做法应该是有后效性的。数据在下面

3 3 30 2
10 10 10
1 1 99
20 20 99

不知道怎么描述这种情况，应该就算是一个二维贪心的反例吧。

但是，如果只能改变列，那么是不会存在后效性的。

同时注意到，n的大小是10，可以暴力2^n，枚举每一行时候变化，然后贪心变列。

#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;
LL a[maxn][maxn];
LL b[maxn][maxn];
int n, m, k;
LL x;
void init() {
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            b[i][j] = a[i][j];
        }
    }
}
int calc(int val) { //复杂度O("1"的个数)
    int ans = 0;
    while (val) {
        val &= val - 1;
        ans++;
    }
    return ans;
}
void change(int val) {
    for (int i = 0; i <= n - 1; ++i) {
        if ((val & (1 << i)) != 0) {
            for (int j = 1; j <= m; ++j) {
                b[i + 1][j] = x;
            }
        }
    }
}
struct node {
    LL sum;
    int id;
    bool operator < (const struct node & rhs) const {
        if (sum != rhs.sum) return sum < rhs.sum;
        else return id < rhs.id;
    }
}c[maxn];
void did(int lef) { //还剩下lef次操作
    int lenc = 0;
    for (int i = 1; i <= m; ++i) {
        LL sum = 0;
        for (int j = 1; j <= n; ++j) {
            sum += b[j][i];
        }
        ++lenc;
        c[lenc].sum = sum;
        c[lenc].id = i;
    }
    sort(c + 1, c + 1 + lenc);
    int en = min(lenc, lef);
    for (int i = 1; i <= en; ++i) {
//        if (c[i].sum >= x * n) return;
        for (int j = 1; j <= n; ++j) {
            b[j][c[i].id] = x;
        }
    }
}
LL now() {
    LL ans = 0;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            ans += b[i][j];
        }
    }
    return ans;
}
void work() {
    cin >> n >> m >> x >> k;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            cin >> a[i][j];
        }
    }
    LL ans = 0;
    int en = (1 << n) - 1;
    for (int i = 0; i <= en; ++i) {
        if (calc(i) > k) continue;
        init();
        change(i);
        did(k - calc(i));
        ans = max(ans, now());
    }
    cout << ans << endl;
}

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

wa的，还要用了BIT的超级复杂的代码

#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;
LL row[maxn][maxn], col[maxn][maxn];
int n, m;
int lowbit(int x) {
    return x & (-x);
}
void upDate(int which, int id, int pos, LL val) {
    if (which == 1) {
        while (pos <= m) {
            row[id][pos] += val;
            pos += lowbit(pos);
        }
    } else {
        while (pos <= n) {
            col[id][pos] += val;
            pos += lowbit(pos);
        }
    }
}
LL ask(int which, int id, int pos) {
    LL ans = 0;
    if (which == 1) {
        while (pos) {
            ans += row[id][pos];
            pos -= lowbit(pos);
        }
    } else {
        while (pos) {
            ans += col[id][pos];
            pos -= lowbit(pos);
        }
    }
    return ans;
}
void tochange(int which, int id, LL val) {
    if (which == 1) {
        for (int i = 1; i <= m; ++i) {
            LL now = ask(which, id, i) - ask(which, id, i - 1);
            upDate(which, id, i, -now);
            upDate(which, id, i, val);
        }
        for (int i = 1; i <= m; ++i) {
            LL now = ask(2, i, id) - ask(2, i, id - 1);
            upDate(2, i, id, -now);
            upDate(2, i, id, val);
        }
    } else {
        for (int i = 1; i <= n; ++i) {
            LL now = ask(which, id, i) - ask(which, id, i - 1);
            upDate(which, id, i, -now);
            upDate(which, id, i, val);
        }
        for (int i = 1; i <= n; ++i) {
            LL now = ask(1, i, id) - ask(1, i, id - 1);
            upDate(1, i, id, -now);
            upDate(1, i, id, val);
        }
    }
}
void work() {
    LL x;
    int k;
    cin >> n >> m >> x >> k;
//    scanf("%d%d%lld%d", &n, &m, &x, &k);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            int t;
            scanf("%d", &t);
            upDate(1, i, j, t);
            upDate(2, j, i, t);
        }
    }
//    tochange(1, 1, 10);
//    cout << ask(2, 2, n) << endl;
    for (int i = 1; i <= k; ++i) {
        LL canadd = 0;
        int id = 0;
        for (int j = 1; j <= n; ++j) {
            LL cango = x * m;
            LL sum = ask(1, j, m);
            if (cango > sum && cango - sum > canadd) {
                canadd = cango - sum;
                id = j;
            }
        }
        for (int j = 1; j <= m; ++j) {
            LL cango = x * n;
            LL sum = ask(2, j, n);
            if (cango > sum && cango - sum >= canadd) {
                canadd = cango - sum;
                id = j + n;
            }
        }
        if (id == 0) break;
        if (id <= n) {
            tochange(1, id, x);
        } else {
            id -= n;
            tochange(2, id, x);
        }
    }
    LL ans = 0;
    for (int i = 1; i <= n; ++i) {
        ans += ask(1, i, m);
    }
    cout << ans << endl;
}

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