Codeforces Round #316 (Div. 2) E

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E Pig and Palindromes

        一个n*m的网格,每个格子有一个字符,你需要从左上角走到右下角(只能往下或往右走),让经过的字符连起来是回文串,问有多少种方案。

        dp。dp(i,j,k),表示从起点走到i行j列的格子,且对称的格子在第k行(列可以计算出来)的方案数,最后的答案是斜线对称轴的方案数的和。但是这样做会超内存,需要滚动数组压第一维变为i%2。


#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
using namespace std;

#define ll long long

const int mod = 1000000007;

char mp[510][510];
int dp[2][510][510];

int n,m;
inline bool verrify(int a,int b,int c,int d){
    if(a<1||a>n||b<1||b>m||c<a||c>n||d<b||d>m)return 0;
    return 1;
}

int main(){
    cin>>n>>m;
    int mid = (n+m+2)>>1;
    for(int i=1;i<=n;i++){
        scanf("%s",mp[i]+1);
    }

    if(mp[1][1]==mp[n][m]){
        dp[1][1][n]=1;
    }

    int ans=0;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m&&(i+j)<=mid;j++){
            if(i==1&&j==1)continue;
            for(int ii=i;ii<=n;ii++){
                int jj=2+n+m-i-j-ii;
                if(jj<j||jj>m)continue;
                dp[i%2][j][ii]=0;
                if(mp[i][j]==mp[ii][jj]){
                    if(verrify(i-1,j,ii+1,jj))dp[i%2][j][ii]+=dp[(i-1)%2][j][ii+1];
                    dp[i%2][j][ii]%=mod;
                    if(verrify(i-1,j,ii,jj+1))dp[i%2][j][ii]+=dp[(i-1)%2][j][ii];
                    dp[i%2][j][ii]%=mod;
                    if(verrify(i,j-1,ii+1,jj))dp[i%2][j][ii]+=dp[i%2][j-1][ii+1];
                    dp[i%2][j][ii]%=mod;
                    if(verrify(i,j-1,ii,jj+1))dp[i%2][j][ii]+=dp[i%2][j-1][ii];
                    dp[i%2][j][ii]%=mod;
                }
                if(i+j==mid){
                    ans+=dp[i%2][j][ii];
                    ans%=mod;
                }
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}


TriCore架构-基于TC397在SOTA过程中AB-SWAP切换底层原理 TC397的SOTA升级机制通过硬件地址重映射实现,利用MMU将逻辑地址动态映射到不同物理地址。其PFlash支持标准/备用地址映射,通常使用PF0+PF1+PF4空间以便于OTA处理。升级时,新固件自动写入非活动分区(A/B分区切换),由地址转换函数完成逻辑地址到物理地址的转换。系统重启后根据寄存器选择启动区域,实现安全无缝的固件更新。 阅读详情

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CodeForces - 570E(dp------------- Codeforces Round #316 (Div. 2)E

题目:给出一个字母矩阵,问从左上角走到右下角的走法有多少种走法能走出回文子串,只能向下或向右走。 思路:正解是O(n^3)的dp,也跑了2.9s,可见这题时限是比较紧的。自然的一个想法是两头同时向中间走,这样可以保证走出的是回文串,问题在于如何dp。一共需要走O(n+m)步,而每一步从左上走过来的和右下走过来的进行匹配,以行数作为下标表示二者即可(因为距离确定的情况下只需要知道一个坐标...

aduqw1129的博客 135

Codeforces Round #316 (Div. 2) C. Replacement

C. Replacementtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputDaniel has a string s, consisting of lowercase English letters and period signs (char

2=10 391

Codeforces Round #316 (Div. 2) D. Tree Requests

Codeforces Round #316 (Div. 2) D. Tree Requests题意: 给一棵树,每个节点上都唯一对应一个单词,给M个询问,问节点u的子树中,深度为h的节点上所有的单词以任意次序组合起来能否构成回文串? 思路: 首先呢,我们必须得处理出内节点所在的深度,而且得知道对应深度下有哪些节点存于vector数组H[d]中,但是问题就是询问仅涉及到节点为u的子树的哪些节点

Step by step 674

Codeforces Round #316 (Div. 2)(ABCD)

A题意: 每行代表一个城市每列的选举人的投票,投票最多的表明该城市支持他,如果支持的人数相同,则选小的,求获得最多城市支持的人。 解析: 直接模拟就好了。 mymy codecode#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int INF = 0x3f3f3f3f;

小G的ACM之路 853

Codeforces Round #316 (Div. 2)D. Tree Requests

链接:http://codeforces.com/contest/570/problem/D 题意:给定一棵n个节点的树(1为根),每个节点上有一个小写字母,m个询问:给定v,h,问在以v为根的子树中深度为h(相对整棵树的深度)的所有点的字符能否组成回文串。 分析:我们来将题目分解一下,首先我们确定构成回文串的条件,很显然要求“字母个数为奇数的字母小于等于1”,这里我们可以用异或储存偶数为0奇

Fsss 506

Codeforces Round #316 (Div. 2)

A - Elections 1 #include<bits/stdc++.h> 2 using namespace std; 3 int n,m,num[105],a[105][105]; 4 int main() 5 { 6 scanf("%d%d",&n,&m); 7 for(int i=1;i<=...

115

[Codeforces Round #316] Replacement

Problem DescriptionDaniel has a string s, consisting of lowercase English letters and period signs (characters ‘.’). Let’s define the operation of replacement as the following sequence of steps: find a

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