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数字三角形模型 AcWing 1018. 最低通行费
2022-07-06 12:15:00 【T_Y_F666】
数字三角形模型 AcWing 1018. 最低通行费
原题链接
算法标签
DP 线性DP
思路
由题意,花费时间应小于(2N−1) , 故不能走回头路, 类似于摘花生问题。
代码
#include<bits/stdc++.h>
#define int long long
#define rep(i, a, b) for(int i=a;i<b;++i)
#define Rep(i, a, b) for(int i=a;i>=b;--i)
using namespace std;
const int N = 105, INF = 0x3f3f3f3f;
int f[N][N], a[N][N];
inline int read(){
int s=0,w=1;
char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
return s*w;
}
void put(int x) {
if(x<0) putchar('-'),x=-x;
if(x>=10) put(x/10);
putchar(x%10^48);
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n=read();
rep(i, 1, n+1){
rep(j, 1, n+1){
a[i][j]=read();
}
}
rep(i, 1, n+1){
rep(j, 1, n+1){
if(i==1&&j==1){
f[i][j]=a[i][j];
}else{
f[i][j]=INF;
if(i>1){
f[i][j]=min(f[i][j], f[i-1][j]+a[i][j]);
}
if(j>1){
f[i][j]=min(f[i][j], f[i][j-1]+a[i][j]);
}
}
}
}
printf("%lld", f[n][n]);
return 0;
}
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