状压DP问题

状态压缩·一

题目传送：#1044 : 状态压缩·一

AC代码：

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n, m, q;
int w[1005];

int dp[1005][1030];//dp[i][j]表示选到第i个位置时j状态能够取得的最大值
int cnt[1030];//代表每一个数的位数上的1的个数

int main() {
    cnt[0] = 0, cnt[1] = 1;
    for(int i = 2; i < 1030; i ++) cnt[i] = cnt[i >> 1] + cnt[i & 1];

    scanf("%d %d %d", &n, &m, &q);
    for(int i = 1; i <= n; i ++) {
        scanf("%d", &w[i]);
    }

    int ans = 0;
    int d = 1 << m;
    for(int i = 1; i <= n; i ++) {
        for(int j = 0; j < (1 << m); j ++) {
            if(cnt[j] <= q) dp[i][j] = max(dp[i-1][j >> 1], dp[i-1][(j >> 1) + (1 << (m - 1))]) + (j & 1) * w[i];
            ans = max(ans, dp[i][j]);
        }
    }

    printf("%d\n", ans);
    return 0;
}

---

Hackers’ Crackdown

题目传送：UVA - 11825 - Hackers’ Crackdown

AC代码：

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = (1 << 16) + 5;
int dp[maxn];//dp[i]表示子集i最多能够分成多少组
int cover[maxn];//cover[i]表示若干集合(i中所表示的集合)的并集

int n, m;
int P[25];//P[i]表示与i相连的数的集合（包含i）

int main() {
    int cas = 1;
    while(scanf("%d", &n) != EOF) {
        if(n == 0) break;
        for(int i = 0; i < n; i ++) {
            scanf("%d", &m);
            P[i] = 1 << i;
            int t;
            while(m --) {
                scanf("%d", &t);
                P[i] |= (1 << t);//并入集合i
            }
        }
        for(int i = 0; i < maxn; i ++) {
            cover[i] = 0;
            for(int j = 0; j < n; j ++) {//if推断j是否在i中，是得话就并入
                if(i & (1 << j)) cover[i] |= P[j];
            }
        }
        dp[0] = 0;
        int tot = (1 << n) - 1;//全集总数
        for(int i = 1; i <= tot; i ++) {//依次枚举全集，由于要先算出前一状态才干推出后一状态
            dp[i] = 0;
            for(int j = i; j; j = (j - 1) & i) {//枚举子集的技巧，重点！
                if(cover[j] == tot) {//i的子集j等于全集，则运行状态转移
                    dp[i] = max(dp[i], dp[i ^ j] + 1);
                }
            }
        }
        printf("Case %d: %d\n", cas ++, dp[tot]);
    }
    return 0;
}

---

Sharing Chocolate

题目传送：UVALive - 4794 - Sharing Chocolate

WF2010的题。

AC代码：

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 16;
int d[1 << maxn][105];
int vis[1 << maxn][105];
int sum[1 << maxn];
int a[maxn];
int n;

int bitcount(int x) {
    return x == 0 ? 0 : bitcount(x / 2) + (x & 1);
}

int dp(int s, int x) {//每次递归找出集合为s。宽为x的巧克力能否够满足要求
    if(vis[s][x]) return d[s][x];
    vis[s][x] = 1;
    int& ans = d[s][x];
    if(bitcount(s) == 1) return ans = 1;//此时为边界。即仅仅有一块巧克力的情况，肯定是满足的
    int y = sum[s] / x;//还有一个边长能够依据这个求得
    for(int s0 = (s - 1) & s; s0; s0 = (s0 - 1) & s) {//枚举子集
        int s1 = s - s0;
        if(sum[s0] % x == 0 && dp(s0, min(x, sum[s0] / x)) && dp(s1, min(x, sum[s1] / x))) return ans = 1;//竖着切（这里假定宽x是竖着的）
        if(sum[s0] % y == 0 && dp(s0, min(y, sum[s0] / y)) && dp(s1, min(y, sum[s1] / y))) return ans = 1;//横着切
    }
    return ans = 0;
}

int main() {
    int cas = 1, x, y;
    while(scanf("%d", &n) != EOF) {
        if(n == 0) break;
        scanf("%d %d", &x, &y);
        for(int i = 0; i < n; i ++) scanf("%d", &a[i]);

        //计算每一个子集的元素的和
        memset(sum, 0, sizeof(sum));
        for(int s = 0; s < (1 << n); s ++) {
            for(int i = 0; i < n; i ++) if(s & (1 << i)) sum[s] += a[i];
        }

        memset(vis, 0, sizeof(vis));
        int ALL = (1 << n) - 1;
        int ans;
        if(sum[ALL] != x * y) ans = 0;
        else ans = dp(ALL, min(x, y));
        printf("Case %d: %s\n", cas ++, ans ?

"Yes" : "No");
    }
    return 0;
}

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