Tower Defence hdu3958
可以转化成求解一条不能接触的最长路径，拐角处可以接触，如果做过channel的话，这道题相对就简单了，不用记录
(x-1, y-1)是否有路径了。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstring>
#include <stack>
#include <cctype>
#include <utility>
#include <map>
#include <string>
#include <climits>
#include <set>
#include <string>
#include <sstream>
#include <utility>
#include <ctime>
using std::priority_queue;
using std::vector;
using std::swap;
using std::stack;
using std::sort;
using std::max;
using std::min;
using std::pair;
using std::map;
using std::string;
using std::cin;
using std::cout;
using std::set;
using std::queue;
using std::string;
using std::stringstream;
using std::make_pair;
using std::getline;
using std::greater;
using std::endl;
using std::multimap;
using std::deque;
using std::random_shuffle;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PAIR;
typedef multimap<int, int> MMAP;
typedef LL TY;
const int MAXN(10010);
const int MAXM(5010);
const int MAXE(10010);
const int MAXK(6);
const int HSIZE(13131);
const int SIGMA_SIZE(26);
const int MAXH(19);
const int INFI((INT_MAX-1) >> 1);
const ULL BASE(31);
const LL LIM(10000000);
const int INV(-10000);
const int MOD(65521);
template<typename T> void checkmax(T &a, T b){if(b > a) a = b;}
template<typename T> void checkmin(T &a, T b){if(b < a) a = b;}
template<typename T> T ABS(const T &a){return a < 0? -a: a;}
int pre[21][10][MAXN], opt[21][10][MAXN];
int MM;
struct HASH_MAP
{
int first[HSIZE], value[MAXN], next[MAXN];
LL state[MAXN];
int size;
void init()
{
memset(first, -1, sizeof(first));
size = 0;
}
void insert(LL ts, int tv, int x, int y, int pid, int op)
{
int h = ts%HSIZE;
for(int i = first[h]; ~i; i = next[i])
if(state[i] == ts)
{
if(tv > value[i])
{
pre[x][y][i] = pid;
opt[x][y][i] = op;
value[i] = tv;
}
return ;
}
checkmax(MM, size);
pre[x][y][size] = pid;
opt[x][y][size] = op;
state[size] = ts;
value[size] = tv;
next[size] = first[h];
first[h] = size++;
}
} hm[2];
HASH_MAP *cur, *last;
int N, M;
int code[11], path[11]; //连通块标号, 是否有路径
int Num[8];
char mp[21][15];
void decode(LL ts)
{
for(int i = 0; i <= M; ++i)
{
code[i] = ts&7;
path[i] = ts&8;
ts >>= 4;
}
}
LL encode()
{
LL ret = 0;
memset(Num, -1, sizeof(Num));
int cnt = 0;
for(int i = M; i >= 0; --i)
if(code[i] == 0)
ret = (ret << 4)|path[i];
else
{
if(Num[code[i]] == -1) Num[code[i]] = ++cnt;
ret = (ret << 4)|Num[code[i]]|path[i];
}
return ret;
}
void updata(int x, int y, int tv, int pid)
{
int lc = (y == 0)? 0: code[y];
int uc = (x == 0)? 0: code[y+1];
int lp = (y == 0)? 0: path[y-1];
int up = (x == 0)? 0: path[y+1];
if(mp[x][y] == 'S' || mp[x][y] == 'T')
{
if(lc == 0 && uc == 0)
{
if(lp || up) return;
if(x < N-1)
{
path[y] = 8;
code[y] = 7;
code[y+1] = 0;
cur->insert(encode(), tv+1, x, y, pid, 1);
}
if(y < M-1)
{
path[y] = 8;
code[y] = 0;
code[y+1] = 7;
cur->insert(encode(), tv+1, x, y, pid, 1);
}
}
else
if(lc == 0 || uc == 0)
{
if(lc)
{
if(up) return;
path[y] = 8;
code[y] = code[y+1] = 0;
cur->insert(encode(), tv+1, x, y, pid, 1);
}
else
{
if(lp) return;
path[y] = 8;
code[y] = code[y+1] = 0;
cur->insert(encode(), tv+1, x, y, pid, 1);
}
}
return;
}
if(mp[x][y] == 'B')
{
if(lc == 0 && uc == 0)
{
path[y] = 0;
code[y] = code[y+1] = 0;
cur->insert(encode(), tv, x, y, pid, 0);
}
return;
}
if(lc == 0 && uc == 0)
{
path[y] = 0;
code[y] = code[y+1] = 0;
cur->insert(encode(), tv, x, y, pid, 0);
if(x == N-1 || y == M-1) return;
if(lp || up) return;
path[y] = 8;
code[y] = code[y+1] = 7;
cur->insert(encode(), tv+1, x, y, pid, 1);
}
else
if(lc == 0 || uc == 0)
{
if(lc)
{
if(up) return;
if(x < N-1)
{
path[y] = 8;
code[y] = lc;
code[y+1] = 0;
cur->insert(encode(), tv+1, x, y, pid, 1);
}
if(y < M-1)
{
path[y] = 8;
code[y] = 0;
code[y+1] = lc;
cur->insert(encode(), tv+1, x, y, pid, 1);
}
}
else
{
if(lp) return;
if(x < N-1)
{
path[y] = 8;
code[y] = uc;
code[y+1] = 0;
cur->insert(encode(), tv+1, x, y, pid, 1);
}
if(y < M-1)
{
path[y] = 8;
code[y] = 0;
code[y+1] = uc;
cur->insert(encode(), tv+1, x, y, pid, 1);
}
}
}
else
if(lc != uc)
{
path[y] = 8;
for(int i = 0; i <= M; ++i)
if(code[i] == uc) code[i] = lc;
code[y] = code[y+1] = 0;
cur->insert(encode(), tv+1, x, y, pid, 1);
}
}
void solve()
{
cur = hm;
last = hm+1;
last->init();
last->insert(0, 0, 0, 0, 0, 0);
for(int i = 0; i < N; ++i)
{
int sz = last->size;
for(int j = 0; j < sz; ++j)
last->state[j] <<= 4;
for(int j = 0; j < M; ++j)
{
cur->init();
sz = last->size;
for(int k = 0; k < sz; ++k)
{
decode(last->state[k]);
updata(i, j, last->value[k], k);
}
swap(cur, last);
}
}
int ans = 0, id;
for(int i = 0; i < last->size; ++i)
{
decode(last->state[i]);
bool flag(true);
for(int j = 0; j <= M; ++j)
if(code[j])
{
flag = false;
break;
}
if(flag)
{
if(last->value[i] > ans)
{
ans = last->value[i];
id = i;
}
}
}
for(int i = 0; i < N; ++i)
for(int j = 0; j < M; ++j)
if(mp[i][j] == '.')
mp[i][j] = 'W';
for(int i = N-1; i >= 0; --i)
for(int j = M-1; j >= 0; --j)
{
if(mp[i][j] == 'W' && opt[i][j][id]) mp[i][j] = '.';
id = pre[i][j][id];
}
printf("%d\n", ans);
for(int i = 0; i < N; ++i)
printf("%s\n", mp[i]);
printf("\n");
}
int main()
{
int TC, n_case(0);
scanf("%d", &TC);
while(TC--)
{
scanf("%d%d", &N, &M);
for(int i = 0; i < N; ++i)
scanf("%s", mp[i]);
printf("Case %d: ", ++n_case);
solve();
}
return 0;
}