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【codeforces】515-D Drazil and Tiles

2015-02-23 21:37 239 查看
先建图,然后把度为1的全丢到队列里面,全跑一遍,没剩下的就可行。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 2005
#define maxm 10005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

char g[maxn][maxn];
int du[maxn * maxn];
queue<pair<int, int> > q;
int n, m, cnt, res;

inline int calc(int i, int j)
{
return i * m + j;
}

void read()
{
memset(g, '#', sizeof g);
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i++) scanf("%s", g[i]);
}

void clear(int i, int j)
{
res++;
du[calc(i, j)] = 0;
if(g[i][j+1] == '.') {
du[calc(i, j+1)]--;
if(du[calc(i, j+1)] == 1) q.push(mp(i, j+1));
}
if(g[i+1][j] == '.') {
du[calc(i+1, j)]--;
if(du[calc(i+1, j)] == 1) q.push(mp(i+1, j));
}
if(i && g[i-1][j] == '.') {
du[calc(i-1, j)]--;
if(du[calc(i-1, j)] == 1) q.push(mp(i-1, j));
}
if(j && g[i][j-1] == '.') {
du[calc(i, j-1)]--;
if(du[calc(i, j-1)] == 1) q.push(mp(i, j-1));
}
}

void work()
{
cnt = res = 0;
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
if(g[i][j] == '.') {
cnt++;
if(g[i][j+1] == '.') du[calc(i, j)]++, du[calc(i, j+1)]++;
if(g[i+1][j] == '.') du[calc(i, j)]++, du[calc(i+1, j)]++;
if(du[calc(i, j)] == 1) q.push(mp(i, j));
}
while(!q.empty()) {
int i = q.front().first, j = q.front().second;
q.pop();
if(g[i][j+1] == '.') clear(i, j+1), g[i][j] = '<', g[i][j+1] = '>';
if(g[i+1][j] == '.') clear(i+1, j), g[i][j] = '^', g[i+1][j] = 'v';
if(i && g[i-1][j] == '.') clear(i-1, j), g[i-1][j] = '^', g[i][j] = 'v';
if(j && g[i][j-1] == '.') clear(i, j-1), g[i][j-1] = '<', g[i][j] = '>';
}
if(cnt == 2 * res) for(int i = 0; i < n; i++) printf("%s\n", g[i]);
else printf("Not unique\n");
}

int main()
{
read();
work();

return 0;
}
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