补一下题,

• c的状态方程都列出来了, 但没想通
• d只剩几个样例没过没验出来,
  
  应该是连通性的问题,刚才换了个写法,都初始化成0x3f过了

• C. 小竹关禁闭
• D 游戏购买!

C. 小竹关禁闭

C. 小竹关禁闭
DP

#include<iostream>
using namespace std;

const int N = 2010;
int dp[N];
int a[N];

inline int fmax(int a, int b) {
    return a < b ? b : a;
}
int main()
{
    int n, k;
    cin >> n >> k;

    for(int i = 1; i <= n; ++i) cin >> a[i];

    for(int i = 1; i <= n; ++i){
        if(i < k+1) dp[i] = fmax(dp[i-1], a[i]);
        else dp[i] = fmax(dp[i-1], dp[i-k-1]+a[i]);
    }

    int res = -1;
    for(int i = 1; i <= n; ++i)
        res = fmax(res, dp[i]);

    cout << res;
    return 0;
}

D 游戏购买!

D 游戏购买!
BFS

#include<iostream>
#include<queue>
#include<cstring>
#define x first
#define y second
using namespace std;

const int N = 2010;
typedef pair<int, int> PII;
int n, m, x;
int sx, sy, ex, ey;
int g[N][N];
int ds[N][N], de[N][N], d[N][N]; // 距离起点, 终点的距离
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

void e_bfs(){
    memset(d, -1, sizeof d);
    queue<PII> q;
    d[ex][ey] = 0;
    q.push({ex, ey});

    while(!q.empty())
    {
        PII t = q.front();
        q.pop();

        for(int i = 0; i < 4; ++i) {
            int nx = t.x+dx[i], ny = t.y+dy[i];
            if(!nx || !ny || nx > n || ny > m || g[nx][ny]==-1) continue;
            //已访问, 跳过
            if(d[nx][ny] != -1) continue;
            q.push({nx, ny});
            d[nx][ny] = d[t.x][t.y]+1;
            if(g[nx][ny] > x) de[nx][ny] = d[nx][ny];
        }
    }
}

long long s_bfs(){
    memset(d, -1, sizeof d);
    queue<PII> q;
    d[sx][sy] = 0;
    q.push({sx, sy});

    long long res = 0x3f3f3f3f;
    while(!q.empty())
    {
        PII t = q.front();
        q.pop();
        if(g[t.x][t.y] > x) res = min(res, (long long)(ds[t.x][t.y]+de[t.x][t.y]));

        for(int i = 0; i < 4; ++i) {
            int nx = t.x+dx[i], ny = t.y+dy[i];
            if(!nx || !ny || nx > n || ny > m || g[nx][ny]==-1) continue;
            //已访问, 跳过
            if(d[nx][ny] != -1) continue;
            q.push({nx, ny});
            d[nx][ny] = d[t.x][t.y]+1;
            if(g[nx][ny] > x) ds[nx][ny] = d[nx][ny];
        }
    }
    return res == 0x3f3f3f3f ? -1 : res;
}
int main()
{
    memset(ds, 0x3f, sizeof ds);
    memset(de, 0x3f, sizeof de);
    scanf("%d%d%d", &n, &m, &x);
    scanf("%d%d%d%d", &sx, &sy, &ex, &ey);
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j)
            scanf("%d", &g[i][j]);

    //终点
    e_bfs();
    printf("%lld", s_bfs());
}