1814: Ural 1519 Formula 1
Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 879 Solved: 328
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Description
Regardless of the fact, that Vologda could not get rights to hold the Winter Olympic games of 20**, it is well-known, that the city will conduct one of the Formula 1 events. Surely, for such an important thing a new race circuit should be built as well as hotels,
restaurants, international airport - everything for Formula 1 fans, who will flood the city soon. But when all the hotels and a half of the restaurants were built, it appeared, that at the site for the future circuit a lot of gophers lived in their holes.
Since we like animals very much, ecologists will never allow to build the race circuit over the holes. So now the mayor is sitting sadly in his office and looking at the map of the circuit with all the holes plotted on it. Problem Who will be smart enough
to draw a plan of the circuit and keep the city from inevitable disgrace? Of course, only true professionals - battle-hardened programmers from the first team of local technical university!.. But our heroes were not looking for easy life and set much more
difficult problem: "Certainly, our mayor will be glad, if we find how many ways of building the circuit are there!" - they said. It should be said, that the circuit in Vologda is going to be rather simple. It will be a rectangle N*M cells in size with a single
circuit segment built through each cell. Each segment should be parallel to one of rectangle's sides, so only right-angled bends may be on the circuit. At the picture below two samples are given for N = M = 4 (gray squares mean gopher holes, and the bold black
line means the race circuit). There are no other ways to build the circuit here. 一個 m * n 的棋盤,有的格子存在障礙,求經過所有非障礙格子的哈密頓迴路個數
Input
The first line contains the integer numbers N and M (2 ≤ N, M ≤ 12). Each of the next N lines contains M characters, which are the corresponding cells of the rectangle. Character "." (full stop) means a cell, where a segment of the race circuit should be built,
and character "*" (asterisk) - a cell, where a gopher hole is located.
Output
You should output the desired number of ways. It is guaranteed, that it does not exceed 2^63-1.
Sample Input
4 4
**..
....
....
....
**..
....
....
....
Sample Output
2
HINT
Source
插頭DP比起其他DP還真的是骨（dai）骼（ma）精（ju）奇（chang）啊... 不過能想到插頭這樣一個很好的比喻還是非常6. cdq的講稿非常清晰， 可惜萬惡的百度文庫需要1下載券... 搞得我花掉了我（b）自（f）己（k）的一個下載券... 這裏附上一個同樣清晰明瞭的講解: 插頭DP——從不會到崩潰. 說着崩潰講的還是很好的嘛~ 但是確實噁心. 省選考到我就gg（flag）. 這道題就是裸題啦~
#include<bits/stdc++.h>
#define clear(a) memset(a, 0, sizeof(a))
using namespace std;
typedef long long lnt;
const int maxn = 2e5 + 5;
char s[15];
lnt ans, dp[2][maxn];
int n, m, c, cnt, ex, ey, tot[2], bit[20];
int mp[15][15], h[maxn], sta[2][maxn];
struct edge{int nxt, v;}e[maxn];
inline void insert(const int &s, const lnt &num) {
int u = s % maxn;
for (int i = h[u]; i; i = e[i].nxt)
if (sta[c][e[i].v] == s) {
dp[c][e[i].v] += num;
return;
}
e[++ cnt].v = ++ tot[c];
e[cnt].nxt = h[u], h[u] = cnt;
sta[c][tot[c]] = s, dp[c][tot[c]] = num;
}
inline void Plug_Dp() {
tot[0] = 1, dp[c][1] = 1;
register int i, j, k, t;
for (i = 1; i <= n; ++ i) {
for (j = 1; j <= tot[c]; ++ j)
sta[c][j] <<= 2;
for (j = 1; j <= m; ++ j) {
clear(h), cnt = 0;
c ^= 1, tot[c] = 0;
for (k = 1; k <= tot[c ^ 1]; ++ k) {
int s = sta[c ^ 1][k];
lnt num = dp[c ^ 1][k];
int p = (s >> bit[j - 1]) & 3;
int q = (s >> bit[j]) & 3;
if (!mp[i][j]) {
if (!p && !q) insert(s, num);
}
else if (!p && !q) {
if (!mp[i + 1][j] || !mp[i][j + 1]) continue;
s += (1 << bit[j - 1]) + (1 << (bit[j] + 1));
//左括號是0， 右括號是1， 四進制在二進制下兩位表示一字節
insert(s, num);
}
else if (!p && q) {
if (mp[i][j + 1]) insert(s, num);
if (mp[i + 1][j]) {
s += (1 << bit[j - 1]) * q - (1 << bit[j]) * q;
insert(s, num);
}
}
else if (p && !q) {
if (mp[i + 1][j]) insert(s, num);
if (mp[i][j + 1]) {
s += (1 << bit[j]) * p - (1 << bit[j - 1]) * p;
insert(s, num);
}
}
else if (p + q == 2) {
int b = 1;
for (t = j + 1; t <= m; ++ t) {
int v = (s >> bit[t]) & 3;
if (v == 1) b ++; //根據括號序列的性質
if (v == 2) b --;
if (!b) {
s -= (1 << bit[t]);
break;
}
}
s -= (1 << bit[j - 1]) + (1 << bit[j]);
insert(s, num);
}
else if (p + q == 4) {
int b = 1;
for (t = j - 2; ~t; -- t) {
int v = (s >> bit[t]) & 3;
if (v == 2) b ++;
if (v == 1) b --;
if (!b) {
s += (1 << bit[t]);
break;
}
}
s -= (1 << (bit[j - 1] + 1)) + (1 << (bit[j] + 1));
insert(s, num);
}
else if (p == 2 && q == 1) {
s -= (1 << (bit[j - 1] + 1)) + (1 << bit[j]);
insert(s, num);
}
else if (p == 1 && q == 2) {
if (i == ex && j == ey) ans += num;
}
}
}
}
printf("%lld\n", ans);
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < 26; ++ i)
bit[i] = i << 1;
for (int i = 1; i <= n; ++ i) {
scanf("%s", s + 1);
for (int j = 1; j <= m; ++ j)
if (s[j] == '.') mp[i][j] = 1, ex = i, ey = j;
}
Plug_Dp();
return 0;
}