poj 3740 Easy Finding 精确匹配

题目链接

dlx的第一题， 真是坎坷.....

 #include <iostream>
 #include <vector>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <map>
 #include <set>
 #include <string>
 #include <queue>
 using namespace std;
 #define pb(x) push_back(x)
 #define ll long long
 #define mk(x, y) make_pair(x, y)
 #define lson l, m, rt<<1
 #define mem(a) memset(a, 0, sizeof(a))
 #define rson m+1, r, rt<<1|1
 #define mem1(a) memset(a, -1, sizeof(a))
 #define mem2(a) memset(a, 0x3f, sizeof(a))
 #define rep(i, a, n) for(int i = a; i<n; i++)
 #define ull unsigned long long
 typedef pair<int, int> pll;
 const double PI = acos(-1.0);
 const double eps = 1e-;
 const int mod = 1e9+;
 const int inf = ;
 const int dir[][] = { {-, }, {, }, {, -}, {, } };
 const int maxn = ;
 const int maxNode = ;
 struct DLX {
     int L[maxNode], R[maxNode], U[maxNode], D[maxNode], row[maxNode], col[maxNode];
     int S[maxn], H[maxn], deep, ans[maxn], sz, n, m;
     void remove(int c) {
         L[R[c]] = L[c] ;
         R[L[c]] = R[c] ;
         for(int i = D[c]; i!=c; i = D[i])
             for(int j = R[i]; j!=i; j = R[j]) {
                 U[D[j]] = U[j] ;
                 D[U[j]] = D[j] ;
                 S[col[j]]--;
             }
     }
     void resume(int c) {
         for(int i = U[c]; i!=c; i = U[i])
             for(int j = L[i]; j!=i; j = L[j]) {
                 S[col[j]]++;
                 D[U[j]] = j;
                 U[D[j]] = j;
             }
         R[L[c]] = c;
         L[R[c]] = c;
     }
     int dfs(int d) {
         if(R[] == ) {
             deep = d;
             return ;
         }
         int c = R[];
         for(int i = R[]; i!=; i = R[i])
             if(S[c]>S[i])
                 c = i;
         remove(c);
         for(int i = D[c]; i!=c; i = D[i]) {
             ans[d] = row[i];
             for(int j = R[i]; j!=i; j = R[j])
                 remove(col[j]);
             if(dfs(d+))
                 return ;
             for(int j = L[i]; j!=i; j = L[j])
                 resume(col[j]);
         }
         resume(c);
         return ;
     }
     void add(int r, int c) {
         sz++;
         row[sz] = r;
         col[sz] = c;
         S[c]++;
         U[sz] = U[c];
         D[sz] = c;
         D[U[c]] = sz;
         U[c] = sz;
         if(~H[r]) {
             R[sz] = H[r];
             L[sz] = L[H[r]];
             L[R[sz]] = sz;
             R[L[sz]] = sz;
         } else {
             H[r] = L[sz] = R[sz] = sz;
         }
     }
     void init(){
         mem1(H);
         for(int i = ; i<=n; i++) {
             R[i] = i+;
             L[i] = i-;
             U[i] = i;
             D[i] = i;
         }
         R[n] = ;
         L[] = n;
         sz = n;
     }
     void solve () {
         init() ;
         for(int i = ; i<=m; i++) {
             for(int j = ; j<=n; j++) {
                 int x;
                 scanf("%d", &x);
                 if(x) {
                     add(i, j);
                 }
             }
         }
         if(dfs()) {
             printf("Yes, I found it\n");
         } else {
             printf("It is impossible\n");
         }
     }
 }dlx;
 int main()
 {
     while(~scanf("%d%d", &dlx.m, &dlx.n)) {
         dlx.solve();
     }
     return ;
 }