POJ 1038 Bugs Integrated, Inc. 三进制压缩DP

黑书经典例题，做法与黑书的思想一样，我这里就不多加解释

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int n, m, k;int g[155][13];int dp[2][180000];int cst[11];int max(int a, int b) { return a > b ? a : b; }int p1[180000], p2[180000], pt1, pt2;bool vis[180000];int f[13];void go(int zt) { //状态解压成数组int i;for(i = 0; i < m; i++) {f[i] = zt%3;zt /=3;}}int to(int *a) { // 返回状态    int ret = 0, i;    for(i = m-1; i >= 0; i--)        ret = ret*3 + a[i];    return ret;}void dfs(int x, int y, int cnt) {if(y >= m) {    int zt = to(f);dp[x&1][zt] = max(dp[x&1][zt], cnt);if(!vis[zt]) p2[pt2++] = zt;vis[zt] = 1;return;}int a = f[y];if(f[y]) f[y]--;dfs(x, y+1, cnt);f[y] = a;if( g[x][y] || g[x-1][y] || g[x][y+1] || g[x-1][y+1]) return;if( x-1 >= 0 && y+2 < m && !g[x][y+2] && !g[x-1][y+2] ) { // height 2if(f[y] <= 1 && f[y+1] <= 1 && f[y+2] <= 1) {    int a = f[y], b = f[y+1], c = f[y+2];    f[y] = f[y+1] = f[y+2] = 2;    dfs(x, y+3, cnt+1);    f[y] = a; f[y+1] = b; f[y+2] = c;}}if( x-2 >= 0 && y+1 < m && !g[x-2][y] && !g[x-2][y+1]) { // height 3if(!f[y] && !f[y+1]) {    f[y] = f[y+1] = 2;dfs(x, y+2, cnt+1);f[y] = f[y+1] = 0;}}}int main() {int i, j, cas;cst[0] = 1;for(i = 1; i <= 11; i++) cst[i] = cst[i-1]*3;scanf("%d", &cas);while( ~scanf("%d%d%d", &n, &m, &k)) {memset(g, 0, sizeof(g));memset(dp, 0, sizeof(dp));for(i = 0; i < k; i++) {int x, y;scanf("%d%d", &x, &y);g[--x][--y] = 1;}pt2 = 0;p2[pt2++] = 0;for(i = 0; i < n-1; i++) {for(j = 0; j < pt2; j++)                p1[j] = p2[j];memset(vis, 0, sizeof(vis));            pt1 = pt2; pt2 = 0;for(j = 0; j < pt1; j++) {    int zt = p1[j];    go(zt);  dfs(i+1, 0, dp[i&1][zt]);}}int ans = 0;for(i = 0; i < pt2; i++)ans = max(ans, dp[(n-1)&1][p2[i]]);printf("%d\n", ans);}return 0;}/*1003 2 0*/