今日SGU 5.18

SGU 125

题意：给你一个数组b[i][j],表示i,j的四周有多少个数字大于它的，问你能不能构造出一个a矩形

收获：dfs  + 剪枝

一行一行的dfs，然后第一行去枚举0-9，下一行判断当前选择能否满足上一行对应列的情况，可以的话就继续dfs

#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define ll long long
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define db double
#define all(a) a.begin(),a.end()
const int mod = 1e9+7;
const int maxn = 1e5+5;
const double eps = 1e-6;
using namespace std;
bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=0,f=1;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while (ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
//inv[1]=1;
//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
int a[6][6],n,b[6][6];
int dx[] = {0,0,1,-1};
int dy[] = {1,-1,0,0};
bool ok(int x,int y){
    int ret = 0;
//    dd(x)dd(y)de(ret)
    rep(i,0,4) if(a[x+dx[i]][y+dy[i]] > a[x][y]) ret++;
    return ret==b[x][y];
}
bool dfs(int x,int y){
//    dd(n)dd(x)de(y)
    int tx = x,ty = y + 1;
    if(x > n){
        rep(i,1,n+1) if(!ok(n,i)) return false;
        rep(i,1,n+1) rep(j,1,n+1) printf("%d%c",a[i][j]," \n"[j==n]);
        return true;
     }
    if(ty > n) ty = 1,tx = x + 1;
    rep(i,0,10){
        a[x][y] = i;
        if(x != 1) if(!ok(x-1,y)) continue;
        if(dfs(tx,ty)) return true;
    }
    return false;
}
int main(){
    scanf("%d",&n);
    rep(i,1,n+1) rep(j,1,n+1) scanf("%d",&b[i][j]);
    if(!dfs(1,1)) puts("NO SOLUTION");
    return 0;
}

原文地址：https://www.cnblogs.com/chinacwj/p/9057472.html