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  • HDU 4804 Campus Design

    HDU 4804

    思路:

    轮廓线dp

    #include<bits/stdc++.h>
    using namespace std;
    #define fi first
    #define se second
    #define pi acos(-1.0)
    #define LL long long
    //#define mp make_pair
    #define pb push_back
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pii pair<int, int>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
    //head
    
    const int N = 12;
    const int MOD = 1e9 + 7;
    int dp[2][1<<N][25];
    int n, m, cur;
    char mp[105][N];
    void update(int a, int b, int t) {
        if(b & (1<<m)) dp[cur][b^(1<<m)][t] += dp[1-cur][a][t-1], dp[cur][b^(1<<m)][t] %= MOD;
    }
    void Update(int a, int b, int t) {
        if(b & (1<<m)) dp[cur][b^(1<<m)][t] += dp[1-cur][a][t], dp[cur][b^(1<<m)][t] %= MOD;
    }
    int main() {
        int c, d;
        while(~scanf("%d%d%d%d", &n, &m, &c, &d)) {
            for (int i = 0; i < n; i++) scanf("%s", mp[i]);
            cur = 0;
            mem(dp, 0);
            dp[cur][(1<<m)-1][0] = 1;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    cur ^= 1;
                    mem(dp[cur], 0);
                    if(mp[i][j] == '1') {
                        for(int k = 0; k < (1<<m); k++) {
                            for (int l = 0; l <= d; l++) {
                                Update(k, (k<<1), l);
                                if(l)update(k, (k<<1)^1, l);
                                if(i && !(k & (1<<m-1))) Update(k, (k<<1)^(1<<m)^1, l);
                                if(j && !(k & 1)) Update(k, (k<<1)^3, l);
                            }
                        }
                    }
                    else {
                        for (int k = 0; k < (1<<m); k++) {
                            for (int l = 0; l <= d; l++) {
                                Update(k, (k<<1)^1, l);
                            }
                        }
                    }
                }
            }
            LL ans = 0;
            for (int i = c; i <= d; i++) ans += dp[cur][(1<<m)-1][i], ans %= MOD;
            printf("%lld
    ", ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/widsom/p/9096060.html
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