hdu 2828 Lamp 重复覆盖

题目链接

给n个灯和m个开关， 每个灯可以由若干个开关控制， 每个开关也可以控制若干个灯， 问你能否找到一种开关的状态， 使得所有的灯都亮。

将灯作为列， 然后把每个开关拆成两行， 开是一行， 关是一行。 然后跑一下就可以。 输出路径的话， 就是用一个vis数组记录一下哪些行被访问过。

#include<bits/stdc++.h>
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 = 1050;
const int maxNode = 20000;
int num;
struct DLX {
    int L[maxNode], R[maxNode], U[maxNode], D[maxNode], row[maxNode], col[maxNode];
    int S[maxn], H[maxn], sz, n, m, vis[maxn*2];
    void remove(int c) {
        for(int i = D[c]; i!=c; i = D[i]) {
            L[R[i]] = L[i];
            R[L[i]] = R[i];
        }
    }
    void resume(int c) {
        for(int i = U[c]; i!=c; i = U[i]) {
            L[R[i]] = i;
            R[L[i]] = i;
        }
    }
    int dfs(int d) {
        if(R[0] == 0) {
            return 1;
        }
        int c = R[0];
        for(int i = R[0]; i!=0; i = R[i])
            if(S[c]>S[i])
                c = i;
        for(int i = D[c]; i!=c; i = D[i]) {
            if(vis[row[i]^1])
                continue;
            remove(i);
            vis[row[i]] = 1;
            for(int j = R[i]; j != i; j = R[j])
                remove(j);
            if(dfs(d+1))
                return 1;
            for(int j = L[i]; j != i; j = L[j])
                resume(j);
            resume(i);
            vis[row[i]] = 0;
        }
        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;
        }
        mem(S);
        R[n] = 0;
        L[0] = n;
        sz = n;
    }
    void solve() {
        init();
        mem(vis);
        int x, y;
        char s[4];
        for(int i = 1; i <= n; i++) {
            scanf("%d", &x);
            while(x--) {
                scanf("%d%s", &y, s);
                if(s[1] == 'N') {
                    add(y<<1, i);
                } else {
                    add(y<<1|1, i);
                }
            }
        }
        if(dfs(0)) {
            for(int i = 1; i <= m-1; i ++) {
                if(vis[i<<1]) {
                    printf("ON ");
                } else {
                    printf("OFF ");
                }
            }
            if(vis[2*m]) {
                puts("ON");
            } else {
                puts("OFF");
            }
        } else {
            puts("-1");
        }
    }
}dlx;
int main()
{
    while(~scanf("%d%d", &dlx.n, &dlx.m)) {
        dlx.solve();
    }
    return 0;
}