poj1038 Bugs Integrated, Inc.

Posted 2020-12-03 wsmrxc

二进制状压练得挺熟， 但是一直没有时间好好研究三进制状压。

于是找了个题目学习了一下， 两天里修修补补总算AC了。

有几个收获。

1. 三进制与十进制的转换的小技巧（利用数组）

2. 刷表法时， 用dfs来更新状态。这种更新状态的方式蛮新颖的，我这种蒟蒻第一次遇见。

3. 还是更新状态。可以先把新状态算出来，再来枚举决策。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 150 + 5;
const int MAXM = 10  + 2;

int N, M, K;
bool badp[MAXN][MAXM];
int p[12] = {1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049};
int f[2][60000];

inline int threetoten(int a[]){
    int tmp = 0;
    for(int i = 0; i < M; i++)
        tmp += a[i] * p[i];
    return tmp;
}

inline void tentothree(int x, int a[]){
    for(int i = 0; i < M; i++)
        a[i] = x % 3, x /= 3;
}

int pre[13], cur[13];
void dfs(int d, int j, int last, int state)
{
    int k;
    f[d][state] = max(f[d][state], last);
    if(j >= M) return ;
    if((j + 1) < M && (pre[j] == 0) && (pre[j + 1] == 0) && (cur[j] == 0) && (cur[j + 1] == 0)){
        cur[j] = cur[j + 1] = 2;
        k = threetoten(cur);
        dfs(d, j + 2, last + 1, k);
        cur[j] = cur[j + 1] = 0;
    }
    if((j + 2) < M && (cur[j] == 0) && (cur[j + 1] == 0) && (cur[j + 2] == 0)){
        cur[j] = cur[j + 1] = cur[j + 2] = 2;
        k = threetoten(cur);
        dfs(d, j + 3, last + 1, k);
        cur[j] = cur[j + 1] = cur[j + 2] = 0;
    }
    dfs(d, j + 1, last, state);
    return ;
}

int solve()
{
    int d = 0; int tmp;
    memset(f[d], -1, sizeof(f[d]));
    for(int i = 0; i < M; i++) pre[i] = (badp[1][i] ? 2 : 1);
    tmp = threetoten(pre);
    f[d][tmp] = 0;

    for(int i = 2; i <= N; i++){
        d ^= 1;
        memset(f[d], -1, sizeof(f[d]));

        for(int j = 0; j < p[M]; j++){
            if(f[d ^ 1][j] == -1) continue;
            tentothree(j, pre);
            for(int k = 0; k < M; k++){
                if(badp[i][k]) cur[k] = 2;
                else cur[k] = (pre[k] == 0 ? 0 : pre[k] - 1);
            }

            tmp = threetoten(cur);
            dfs(d, 0, f[d ^ 1][j], tmp);
        }
    }
    int ans = 0;
    for(int i = 0; i < p[M]; i++) ans = max(ans, f[d][i]);
    return ans;
}

int main()
{
    int T; cin>>T;
    while(T--)
    {
        cin>>N>>M>>K;
        memset(badp, false, sizeof(badp));
        for(int i = 1, x, y; i <= K; i++){
            scanf("%d%d", &x, &y);
            badp[x][y - 1] = true;
        }
        printf("%d
", solve());
    }
    return 0;
}