codefore 213 C Relay Race （朴素DP）

原题地址：http://codeforces.com/problemset/problem/213/C

题意：略

题解：dp,状态为点的位置，直接枚举复杂度是n4,利用两个点坐标之和之间的关系，优化到n3。

#include<bits/stdc++.h>

#define clr(x,y) memset((x),(y),sizeof(x))

using namespace std;
typedef long long LL;

const int maxn=300;
const int INF=1e8;

int A[maxn+5][maxn+5];
int dp[maxn+5][maxn+5];
int dx[]={-1,-1,0,0};
int dx2[]={-1,0,-1,0};

int n;

int main(void)
{
    #ifdef ex
    freopen ("../in.txt","r",stdin);
    //freopen ("../out.txt","w",stdout);
    #endif

    scanf("%d",&n);
    for (int i=1;i<=n;++i)
    {
        for (int j=1;j<=n;++j)
        {
            scanf("%d",&A[i][j]);
        }
    }

    int x1,y1,x2,y2,tmp;

    //cout<<dp[1][1][1]<<‘ ‘<<dp[2][1][1]<<endl;
    for (int i=1;i<=n;++i)
        for (int j=1;j<=n;++j)
            dp[i][j]=-INF;

    dp[1][1]=A[1][1];
    for (int i=3;i<=2*n;++i)
    {
        for (int j=min(i-1,n);j>=1;--j)
        {
            for (int k=min(i-1,n);k>=1;--k)
            {
                int T=-INF;
                for (int q=0;q<=3;++q)
                {
                    x1=j+dx[q];
                    x2=k+dx2[q];
                    y1=i-1-x1;
                    y2=i-1-x2;

                    if (x1<=0 || x1>n || y1<=0 || y1>n) continue;
                    if (x2<=0 || x2>n || y2<=0 || y2>n) continue;

                    if (j==k) tmp=A[j][i-j];
                    else tmp=A[j][i-j]+A[k][i-k];

                    T=max(T,dp[x1][x2]+tmp);
                }
                dp[j][k]=T;
                //printf("%d %d %d %d\n",i,j,k,dp[j][k]);
            }
        }
    }

    int ans=dp[n][n];
    printf("%d\n",ans);
}