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+7; const int maxn = 1e3+6; 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[maxn][maxn],b[maxn][maxn]; int main(){ int n,m; scanf("%d%d",&n,&m); rep(i,0,n+2) 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,1,n+1) rep(j,1,m+1) scanf("%d",&a[i][j]),b[i][j]=a[i][j]; repd(j,m,1) repd(i,n,1){ b[i][j]=min(b[i][j],b[i+1][j]); b[i][j]=min(b[i][j],b[i-1][j+1]); b[i][j]=min(b[i][j],b[i][j+1]);//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,1,n+1)rep(j,1,m+1) printf("%d%c",b[i][j]," \n"[j==m]); return 0; }