【洛谷1130】红牌

思路：

f[i][j]表示前i步在第j个小组的最短天数

转移：f[i][j]=min(f[i-1][j],f[i-1][x])+f[i][j]

注：当j==1时，x=m；else x=j-1

#include<cstdio>
#include<cstring>
using namespace std;
long long c[1001][2001],f[1001][2001];
long long min(long long,long long);
int main(){
    int h,l;
    long long ans=10000000000;
    scanf("%d %d",&l,&h);
    for (int i=1;i<=h;i++)
        for (int j=1;j<=l;j++)
            scanf("%d",&c[i][j]);
    for (int i=1;i<=l;i++)
        for (int j=1;j<=h;j++){
            if (j==1)
                f[j][i]=min(f[h][i-1],f[j][i-1])+c[j][i];
            else 
                f[j][i]=min(f[j-1][i-1],f[j][i-1])+c[j][i];
        }
    for (int i=1;i<=h;i++)
       ans=min(ans,f[i][l]);
    printf("%lld",ans);
    return 0;
}
long long min(long long x,long long y){
    return x<y?x:y;
}