Gym

题目传送门

题目大意：给出一幅海洋的描述，0为海平面，负数即有水，在给出的xy坐标的底部安放抽水机，问最多能有多少水。水往低处流，且八个方向都可以。

思路：bfs，记录到每一个节点有效的最低海平面，然后尝试更新周围的点。

但这道题需要优先队列，有效海平面最低的先出队，否则会TLE。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string.h>
#include<sstream>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<bitset>
#define CLR(a,b) memset((a),(b),sizeof((a))) 
using namespace std;
typedef long long ll;
inline int rd() {
    int f = 1; int x = 0; char s = getchar();
    while (s<'0' || s>'9') { if (s == '-')f = -1; s = getchar(); }
    while (s >= '0'&&s <= '9') { x = x * 10 + s - '0'; s = getchar(); }x *= f;
    return x;
}
const int maxn = 510;
ll mp[maxn][maxn],mmp[maxn][maxn],vis[maxn][maxn];
int n, m, x, y;
int fx[8][2] = { {-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0} };
struct dian {
    int x, y;
    ll low;
    inline dian(){}
    inline dian(int x,int y,ll low):x(x),y(y),low(low){}
    inline friend bool operator <(const dian & a, const dian &b) {
        return a.low > b.low;
    }
};
inline bool check(dian &s)
{
    if (s.x <= 0 || s.y <= 0 || s.x > n || s.y > m||vis[s.x][s.y])return false;
    if (mp[s.x][s.y] > 0)return false;
    return true;
}
priority_queue<dian >q;
int main() {
    scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                scanf("%I64d", &mp[i][j]);
            }
    }
        scanf("%d%d", &x, &y);
        if (mp[x][y] > 0) {
            printf("0
");
            return 0;
        }
        q.push(dian(x, y,mp[x][y]));
        ll ans =0;
        mmp[x][y] = mp[x][y];
        vis[x][y] = 1;
        while (!q. empty()) {
            dian st = q.top();
            q.pop();
            for (int i = 0; i < 8; i++)
            {
                dian now = st;
                now.x += fx[i][0], now.y += fx[i][1];
                if (!check(now))continue;
                now.low= -min(-st.low, -mp[now.x][now.y]);
                if (now.low < mmp[now.x][now.y]) {
                    q.push(now);
                    mmp[now.x][now.y] = now.low;
                    vis[now.x][now.y] = 1;
                }
            }
        }
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                ans -= mmp[i][j];
            }
        }
        printf("%I64d
", ans);

}