D. Nash Matrix

题目链接：http://codeforces.com/contest/1316/problem/D

题目意思太长了我就不多说了

想法：

走到自己位置上的一定是'X'，然后同一路径上的目的地一定一样，可以从每个‘X’开始把同一目的地的都搜过去，处理出来，然后-1的情况就是所有联通的-1构成一个死循环就好了，如果最后发现有没有被处理到的，说明无法构造出来。

#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <math.h>
#include <cstdio>
#include <iomanip>
#include <time.h>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>

#define LL long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back

const double eps = 1e-10;
const int maxn = 1e3 + 10;
const LL mod = 1e9 + 7;
const LL INF = 1e18;

int sgn(double a){return a < -eps ? -1 : a < eps ? 0 : 1;}
using namespace std;

pii p[maxn][maxn];
char m[maxn][maxn];
int n;

inline void dfs1(int x,int y,int xx,int yy) {
    if (x > 1 && p[x-1][y].first == xx && p[x-1][y].second == yy && m[x-1][y] == '#') {
        m[x - 1][y] = 'D';
        dfs1(x-1,y,xx,yy);
    }
    if (x < n && p[x+1][y].first == xx && p[x+1][y].second == yy && m[x+1][y] == '#') {
        m[x + 1][y] = 'U';
        dfs1(x+1,y,xx,yy);
    }
    if (y > 1 && p[x][y-1].first == xx && p[x][y-1].second == yy && m[x][y-1] == '#') {
        m[x][y-1] = 'R';
        dfs1(x,y-1,xx,yy);
    }
    if (y < n && p[x][y+1].first == xx && p[x][y+1].second == yy && m[x][y+1] == '#') {
        m[x][y+1] = 'L';
        dfs1(x,y+1,xx,yy);
    }
}


inline void dfs2(int x,int y) {
    if (x > 1 && m[x-1][y] == '?') {
        m[x-1][y] = 'D';
        dfs2(x-1,y);
    }
    if (x < n && m[x+1][y] == '?') {
        m[x+1][y] = 'U';
        dfs2(x+1,y);
    }
    if (y > 1 && m[x][y-1] == '?') {
        m[x][y-1] = 'R';
        dfs2(x,y-1);
    }
    if (y < n && m[x][y+1] == '?') {
        m[x][y+1] = 'L';
        dfs2(x,y+1);
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1;i <= n;i++) {
        for (int j = 1;j <= n;j++)
            m[i][j] = '#';
    }
    for (int i = 1;i <= n;i++) {
        for (int j = 1;j <= n;j++) {
            cin >> p[i][j].first >> p[i][j].second;
            if (i == p[i][j].first && j == p[i][j].second)
                m[i][j] = 'X';
            if (-1 == p[i][j].first && -1 == p[i][j].second)
                m[i][j] = '?';
        }
    }
    for (int i = 1;i <= n;i++) {
        for (int j = 1;j <= n;j++) {
            if (m[i][j] == 'X') {
                dfs1(i,j,i,j);
            }
        }
    }
    for (int i = 1;i <= n;i++) {
        for (int j = 1;j <= n;j++) {
            if (m[i][j] == '?') {
                if (m[i+1][j] == '?' && i < n) {
                    m[i][j] = 'D';
                    m[i+1][j] = 'U';
                    dfs2(i,j);
                    dfs2(i+1,j);
                }
                if (m[i][j+1] == '?' && j < n) {
                    m[i][j] = 'R';
                    m[i][j+1] = 'L';
                    dfs2(i,j);
                    dfs2(i,j+1);
                }
            }
        }
    }
    for(int i = 1;i <= n;i++)
        for(int j = 1;j <= n;j++) {
            if(m[i][j]=='#'||m[i][j]=='?'){cout<<"INVALID";return 0;}
        }
    cout<<"VALID"<<endl;
    for(int i = 1;i <= n;i++) {
        for(int j = 1;j <= n;j++)
            cout << m[i][j];
        cout<<endl;
    }
    return 0;
}