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-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 305;
const int maxNode = 5000;
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[0] == 0) {
            deep = d;
            return 1;
        }
        int c = R[0];
        for(int i = R[0]; i!=0; 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+1))
                return 1;
            for(int j = L[i]; j!=i; j = L[j])
                resume(col[j]);
        }
        resume(c);
        return 0;
    }
    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 = 0; i<=n; i++) {
            R[i] = i+1;
            L[i] = i-1;
            U[i] = i;
            D[i] = i;
        }
        R[n] = 0;
        L[0] = n;
        sz = n;
    }
    void solve () {
        init() ;
        for(int i = 1; i<=m; i++) {
            for(int j = 1; j<=n; j++) {
                int x;
                scanf("%d", &x);
                if(x) {
                    add(i, j);
                }
            }
        }
        if(dfs(0)) {
            printf("Yes, I found it
");
        } else {
            printf("It is impossible
");
        }
    }
}dlx;
int main()
{
    while(~scanf("%d%d", &dlx.m, &dlx.n)) {
        dlx.solve();
    }
    return 0;
}