E. Marbles 状压dp

E. Marbles

这个是一个状压dp

题目大意是：给你一个数组，数组的数在1到20之间，有一个操作就是交换相邻的两个数，问 让所有相同的数相邻的最小操作次数

dp[s] 表示s状态下的操作次数，w[i][j] 表示 i 前面  j  的个数，c[i] 表示一开始使 i 放到最前面的最小操作次数。

val[s][i] 表示在s状态下，将 i 放到前面的操作次数。

那么dp[tmp]=min(dp[tmp],dp[s]+val[s][j])

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <string>
#include <algorithm>
#include <iostream>
#include <map>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
const int maxn = 4e5 + 10;
typedef long long ll;
vector<int>num;
ll dp[1 << 20], c[21];
ll w[21][21], val[1 << 20][21];
int vis[21], b[21], a[maxn], sum[21];
bool vit[21];

void judge(int x) {
    num.clear();
    for (int i = 1; x; i++) {
        if (x & 1) num.push_back(i);
        x >>= 1;
    }
}

int main() {
    int n, len = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        if (!vit[a[i]]) b[++len] = a[i];
        vit[a[i]] = 1;
    }
    sort(b + 1, b + 1 + len);
    for (int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + 1 + len, a[i]) - b;
    for (int i = 1; i <= n; i++) {
        sum[a[i]]++;
        for (int j = 1; j <= len; j++) {
            if (j == a[i]) continue;
            w[a[i]][j] += sum[j];
        }
    }
    memset(dp, inf64, sizeof(dp));
    for (int i = 1; i <= n; i++) {
        c[a[i]] += i - 1 - vis[a[i]];
        vis[a[i]]++;
    }
    for (int i = 1; i <= len; i++) dp[1 << (i - 1)] = c[i];
    for (int i = 0; i < (1 << len); i++) {
        judge(i);
        for (int j = 1; j <= len; j++) {
            ll x = c[j];
            int tmp = 1 << (j - 1);
            if ((tmp | i) == i) continue;
            for (int k = 0; k < num.size(); k++) x -= w[j][num[k]];
            val[i][j] = x;
        }
    }
    for (int i = 0; i < (1 << len); i++) {
        for (int j = 1; j <= len; j++) {
            int tmp1 = 1 << (j - 1);
            if ((tmp1 | i) == i) continue;
            if (dp[i] >= inf64) continue;
            tmp1 |= i;
            dp[tmp1] = min(dp[tmp1], dp[i] + val[i][j]);
        }
    }
    printf("%lld\n", dp[(1 << len) - 1]);
    return 0;
}

View Code