今日SGU 5.10

SGU 168

题意：从a矩阵求出b矩阵，规则直接看题目就行了，不好打字说明

收获：dp

#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+;
const int maxn = 1e3+;
const double eps = 1e-;
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==?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=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=,f=;char ch=getchar();
    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while (ch>=''&&ch<=''){x=x*+ch-'';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[maxn][maxn],b[maxn][maxn];
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    rep(i,,n+) mt(b[i],inf);
//    rep(i,0,n+2)rep(j,0,m+2) printf("%d%c",b[i][j]," \n"[j-1==m]);
    rep(i,,n+) rep(j,,m+) scanf("%d",&a[i][j]),b[i][j]=a[i][j];
    repd(j,m,) repd(i,n,){
        b[i][j]=min(b[i][j],b[i+][j]);
        b[i][j]=min(b[i][j],b[i-][j+]);
        b[i][j]=min(b[i][j],b[i][j+]);//比如6 5 1，这个矩阵你不加这句就是错的，因为它这个题目比较坑只要求y，然后x可以越界，越界的默认无穷大
    }
    rep(i,,n+)rep(j,,m+) printf("%d%c",b[i][j]," \n"[j==m]);
    return ;
}