[LA3620]Manhattan Wiring
[LA3620]Manhattan Wiring
试题描述
![](https://images2015.cnblogs.com/blog/822137/201706/822137-20170607220948325-664866711.png)
输入
![](https://images2015.cnblogs.com/blog/822137/201706/822137-20170607221023997-986728304.png)
输出
![](https://images2015.cnblogs.com/blog/822137/201706/822137-20170607221053872-857177828.png)
输入示例
5 5 0 0 0 0 0 0 0 0 3 0 2 0 2 0 0 1 0 1 1 1 0 0 0 0 3 2 3 2 2 0 0 3 3 6 5 2 0 0 0 0 0 3 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 2 3 0 5 9 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 9 9 3 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 3 9 9 0 0 0 1 0 0 0 0 0 0 2 0 1 0 0 0 0 3 0 0 0 1 0 0 0 0 2 0 0 0 1 0 0 0 0 3 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 9 9 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 2 0 0
输出示例
18 2 17 12 0 52 43
数据规模及约定
见“输入”
题解
我们把“连线”的过程改为“铺地砖”的过程,总共有 11 种地砖,每种地砖上的图案连接了两个不同的边界,或只触碰了一个边界,或没有图案,具体见下图:
其中,有障碍的格子只能铺 0 号砖,有数字 2 或 3 的格子只能铺 1 到 4 号砖,空地可以铺 0 或 5 到 10 号砖。
然后我们就可以轮廓线 dp 了,把状态表示成上一行的底部是否有线,这一行的底部是否有线,当前格子的左边是否有线,具体见下图:
带绿点的格子表示当前格子。那么上图的状态就是 (02000100)3 了(我习惯先读上面一行,再读下面一行,最后读竖直边上的数字),注意这里 2 连出的线与 3 连出的线进行了区分,因为不能让 2 和 3 连到一起。
转移的时候需要判断一些不合法情况:线头接到了没有线头和它相连的地方,不同类型线头接在了一起,或是有一个线头等你去接而你没有理它。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <algorithm> using namespace std; const int BufferSize = 1 << 16; char buffer[BufferSize], *Head, *Tail; inline char Getchar() { if(Head == Tail) { int l = fread(buffer, 1, BufferSize, stdin); Tail = (Head = buffer) + l; } return *Head++; } int read() { int x = 0, f = 1; char c = Getchar(); while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); } while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); } return x * f; } #define maxn 15 #define maxs 59060 #define maxb 11 #define oo 2147483647 struct Blo { bool L, U, R, D; int v; Blo() {} Blo(bool _l, bool _u, bool _r, bool _d, int _v): L(_l), U(_u), R(_r), D(_d), v(_v) {} } bls[maxb]; int n, m, Map[maxn][maxn], f[maxn][maxn][maxs], tri[maxn]; void up(int& a, int b) { a = min(a, b); return ; } char str[maxn]; char* tri_(int x) { int l = 0; while(x) str[l++] = x % 3 + '0', x /= 3; while(l <= m) str[l++] = '0'; str[l] = 0; return str; } int main() { bls[0] = Blo(0, 0, 0, 0, 0); bls[1] = Blo(1, 0, 0, 0, 1); bls[2] = Blo(0, 1, 0, 0, 1); bls[3] = Blo(0, 0, 1, 0, 1); bls[4] = Blo(0, 0, 0, 1, 1); bls[5] = Blo(1, 1, 0, 0, 2); bls[6] = Blo(1, 0, 1, 0, 2); bls[7] = Blo(1, 0, 0, 1, 2); bls[8] = Blo(0, 1, 1, 0, 2); bls[9] = Blo(0, 1, 0, 1, 2); bls[10] = Blo(0, 0, 1, 1, 2); tri[0] = 1; for(int i = 1; i < maxn; i++) tri[i] = tri[i-1] * 3; while(1) { n = read(); m = read(); if(!n && !m) break; for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) Map[i][j] = read(); int all = tri[m+1] - 1; for(int i = 1; i <= n + 1; i++) for(int j = 1; j <= m; j++) for(int S = 0; S <= all; S++) f[i][j][S] = oo; f[1][1][0] = 0; for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) for(int S = 0; S <= all; S++) if(f[i][j][S] < oo) { // printf("%d %d %s: %d\n", i, j, tri_(S), f[i][j][S]); if(Map[i][j] == 1) { if(S % 3 || S / tri[m]) continue; if(j < m) up(f[i][j+1][S/3%tri[m-1]], f[i][j][S] + bls[0].v); else up(f[i+1][1][S/3%tri[m-1]], f[i][j][S] + bls[0].v); } if(Map[i][j] == 2) { for(int c = 1; c <= 4; c++) { if(S % 3 > 0 ^ bls[c].U > 0) continue; if(S % 3 && S % 3 != 1) continue; if(S / tri[m] > 0 ^ bls[c].L > 0) continue; if(S / tri[m] && S / tri[m] != 1) continue; if(j == m && bls[c].R) continue; int tS = S / 3 % tri[m-1] + (int)bls[c].D * tri[m-1] + (int)bls[c].R * tri[m]; if(j < m) up(f[i][j+1][tS], f[i][j][S] + bls[c].v); else up(f[i+1][1][tS], f[i][j][S] + bls[c].v); } } if(Map[i][j] == 3) { for(int c = 1; c <= 4; c++) { if(S % 3 > 0 ^ bls[c].U > 0) continue; if(S % 3 && S % 3 != 2) continue; if(S / tri[m] > 0 ^ bls[c].L > 0) continue; if(S / tri[m] && S / tri[m] != 2) continue; if(j == m && bls[c].R) continue; int tS = S / 3 % tri[m-1] + (int)bls[c].D * 2 * tri[m-1] + (int)bls[c].R * 2 * tri[m]; if(j < m) up(f[i][j+1][tS], f[i][j][S] + bls[c].v); else up(f[i+1][1][tS], f[i][j][S] + bls[c].v); } } if(!Map[i][j]) { for(int c = 0; c <= 10; c ? c++ : (c = 5)) { int tp = 0; if(S % 3 > 0 ^ bls[c].U > 0) continue; if(S % 3) tp = S % 3; if(S / tri[m] > 0 ^ bls[c].L > 0) continue; if(S / tri[m] && tp && S / tri[m] != tp) continue; if(S / tri[m]) tp = S / tri[m]; if(j == m && bls[c].R) continue; if(tp) { int tS = S / 3 % tri[m-1] + (int)bls[c].D * tp * tri[m-1] + (int)bls[c].R * tp * tri[m]; if(j < m) up(f[i][j+1][tS], f[i][j][S] + bls[c].v); else up(f[i+1][1][tS], f[i][j][S] + bls[c].v); } else for(tp = 1; tp <= 2; tp++) { int tS = S / 3 % tri[m-1] + (int)bls[c].D * tp * tri[m-1] + (int)bls[c].R * tp * tri[m]; if(j < m) up(f[i][j+1][tS], f[i][j][S] + bls[c].v); else up(f[i+1][1][tS], f[i][j][S] + bls[c].v); } } } } printf("%d\n", f[n+1][1][0] < oo ? (f[n+1][1][0] >> 1) : 0); } return 0; }
代码贼难写。。。QAQ