SRM 616

来源：互联网发布：淘宝上哪个冒菜好吃编辑：程序博客网时间：2024/05/16 09:49

首先枚举每个L的拐点是哪一行，n^3

然后一行dp过去，每个L有三种状态：还没放，放下在持续中（即L的横还没结束），已经放完了，用3^3表示这个状态

dp[m][3^3]，然后转移枚举2^3，看每个L的状态是否要变成下一个状态，除了0->1这种状态转变，其他状态转变方案数都是乘以1

0->1就是看这个位置向上最多能延伸多少（即L的竖线，要看地图，还有看其他的L）

唔，这方法复杂度有点高：n^3*m*27*8，不过最大的数据600ms左右也跑出来了。。。。。（不像正解啊T T

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <cstring>using namespace std;class ThreeLLogo {public:long long countWays(vector<string> );};long long dp[35][27];int up[33][33];long long ThreeLLogo::countWays(vector<string> s) {int i, j, k;int a, b, c;int n, m;int t1, t2, t3, tt1, tt2, tt3;n = s.size();m = s[0].size();long long ans = 0;for (i = 0; i < n; ++i) {for (j = 0; j < m; ++j) {if (s[i][j] == '#')up[i][j] = -1;else if (i == 0)up[i][j] = 0;elseup[i][j] = up[i - 1][j] + 1;}}for (a = 1; a < n; ++a) {for (b = a; b < n; ++b) {for (c = b; c < n; ++c) {memset(dp, 0, sizeof(dp));dp[0][0] = 1;for (i = 0; i < m; ++i) {for (j = 0; j < 27; ++j) {if (dp[i][j] == 0)continue;t1 = j % 3;t2 = j / 3 % 3;t3 = j / 9 % 3;if (t1 == 1 && s[a][i] == '#')continue;if (t2 == 1 && s[b][i] == '#')continue;if (t3 == 1 && s[c][i] == '#')continue;if (a == b && t1 < t2)continue;if (a == c && t1 < t3)continue;if (b == c && t2 < t3)continue;for (k = 0; k < 8; ++k) {long long cost = 1;tt1 = t1 + k % 2;tt2 = t2 + k / 2 % 2;tt3 = t3 + k / 4;if (tt1 == 1 && s[a][i] == '#')continue;if (tt2 == 1 && s[b][i] == '#')continue;if (tt3 == 1 && s[c][i] == '#')continue;if (tt1 > 2 || tt2 > 2 || tt3 > 2)continue;if (t1 == 0 && tt1 == 1) {cost *= up[a][i];}if (t2 == 0 && tt2 == 1) {long long t;t = up[b][i];if (t1 == 1 || tt1 == 1)t = min(t, (long long) b - a - 1);if (t < 0)t = 0;cost *= t;}if (t3 == 0 && tt3 == 1) {long long t;t = up[c][i];if (t1 == 1 || tt1 == 1)t = min(t, (long long) c - a - 1);if (t2 == 1 || tt2 == 1)t = min(t, (long long) c - b - 1);if (t < 0)t = 0;cost *= t;}int newj = tt1 + tt2 * 3 + tt3 * 9;dp[i + 1][newj] += dp[i][j] * cost;}}}ans += dp[m][26];}}}return ans;}

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