題目鏈接

hdu1693

題解

插頭\(dp\)

特點：範圍小，網格圖，連通性
輪廓線：已決策點和未決策點的分界線
插頭：存在於網格之間，表示著網格建的信息，此題中表示兩個網格間是否連邊
狀態表示：當前點\((i,j)\)和輪廓線上\(m + 1\)個插頭的狀態

狀態轉移：

我們用\(f[i][j][s]\)表示如上的狀態，最後一次決策點為\((i,j)\)，輪廓線上插頭狀態為\(s\)的方案數
比如上圖\(s = 1101001\)

之後我們擴展新的點，枚舉它插頭的狀態進行轉移

在本題中，要使最終形成若幹回路，每個點度數必須為\(2\)，所以我們擴展點的時候記錄它已有的插頭數，然後剩余的插頭數就可以唯一確定

然後就可以\(O(nm2^m)\)過了這道插頭\(dp\)入門題

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
 

#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 12,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while
 
 (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int n,m,S[maxn][maxn];
LL f[maxn][maxn][1 << maxn];
void work(int C){
    cls(f);
    if (!S[1][1]) f[1][1][0] = 1;
    else {
        if (S[1][2] == 0 || S[2][1] == 0){
            printf("Case %d: There are 0 ways to eat the trees.\n",C);
            return;
        }
        f[1][1][3] = 1;
    }
    int maxv = (1 << m + 1) - 1,cnt,e,t;
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= m; j++){
            if (i == n && j == m) break;
            for (int s = 0; s <= maxv; s++){
                if (!f[i][j][s]) continue;
                if (j == m){
                    if (s & 1){
                        if (i + 2 <= n && S[i + 2][1])
                            f[i + 1][1][(s >> 1) << 2 | 1] += f[i][j][s];
                        if (S[i + 1][2])
                            f[i + 1][1][(s >> 1) << 2 | 2] += f[i][j][s];
                    }
                    else {
                        if (!S[i + 1][1]) f[i + 1][1][(s >> 1) << 2] += f[i][j][s];
                        else {
                            if (i + 2 > n || !S[i + 2][1] || !S[i + 1][2]) continue;
                            f[i + 1][1][(s >> 1) << 2 | 3] += f[i][j][s];
                        }
                    }
                }
                else {
                    cnt = ((s >> j) & 1) + ((s >> j + 1) & 1);
                    t = (s >> j) & 3; e = s ^ (t << j);
                    if (cnt && !S[i][j + 1]) continue;
                    if (cnt == 2) f[i][j + 1][e] += f[i][j][s];
                    else if (cnt == 1){
                        if (i + 1 <= n && S[i + 1][j + 1])
                            f[i][j + 1][e | (1 << j)] += f[i][j][s];
                        if (j + 2 <= m && S[i][j + 2])
                            f[i][j + 1][e | (1 << j + 1)] += f[i][j][s];
                    }
                    else {
                        if (!S[i][j + 1]) f[i][j + 1][e] += f[i][j][s];
                        else {
                            if (i + 1 > n || j + 2 > m || !S[i + 1][j + 1] || !S[i][j + 2])
                                continue;
                            f[i][j + 1][e | (3 << j)] += f[i][j][s];
                        }
                    }
                }
            }
        }
    }
    LL ans = 0;
    for (int s = 0; s <= maxv; s++)
        ans += f[n][m][s];
    printf("Case %d: There are %lld ways to eat the trees.\n",C,ans);
}
int main(){
    int T = read();
    REP(t,T){
        n = read(); m = read();
        REP(i,n) REP(j,m) S[i][j] = read();
        if (n == 1 || m == 1){
            if (!S[n][m]) printf("Case %d: There are 1 ways to eat the trees.\n",t);
            else printf("Case %d: There are 0 ways to eat the trees.\n",t);
            continue;
        }
        work(t);
    }
    return 0;
}

hdu1693 Eat the Trees 【插頭dp】