从外面一点一点往里面拓展(floodfill)，每次找出最小的一个点，计算它对答案的贡献就好了。。。

找最小的点的话，直接pq就行

 

1 /************************************************************** 2     Problem: 1736 3     User: rausen 4     Language: C++ 5     Result: Accepted 6     Time:196 ms 7     Memory:2116 kb 8 ****************************************************************/ 9  10 #include 
      
       11 #include 
       
        12  13 using namespace std;14 typedef long long ll;15 const int N = 305;16 const int dx[] = {
    0, 0, 1, -1};17 const int dy[] = {
    1, -1, 0, 0};18  19 inline int read();20  21 struct data {22     int x, y, h;23     data(int _x = 0, int _y = 0, int _h = 0) : x(_x), y(_y), h(_h) {}24      25     inline bool operator < (const data &d) const {   26         return h > d.h;27     }28 };29  30 int n, m;31 int mp[N][N], v[N][N];32 priority_queue 
         h;33  34 inline ll work(){35     ll res = 0;36     int x, y, k;37     data now;38     while (!h.empty()) {39         now = h.top(), h.pop();40         for (k = 0; k < 4; ++k) {41             x = now.x + dx[k], y = now.y + dy[k];42             if (x <= 0 || y <= 0 || x > n || y > m || v[x][y]) continue;43             v[x][y] = 1;44             if (mp[x][y] < now.h)45                 res += now.h - mp[x][y], mp[x][y] = now.h;46             h.push(data(x, y, mp[x][y]));47         }48     }49     return res;50 }51  52 int main() {53     int i, j;54     m = read(), n = read();55     for (i = 1; i <= n; ++i)56         for (j = 1; j <= m; ++j) mp[i][j] = read();57     for (i = 1; i <= n; ++i)58         for (j = 1; j <= m; ++j)59             if (i == 1 || j == 1 || i == n || j == m)60                 h.push(data(i, j, mp[i][j])), v[i][j] = 1;61     printf("%lld\n", work());62     return 0;63 }64  65 inline int read() {66     static int x;67     static char ch;68     x = 0, ch = getchar();69     while (ch < '0' || '9' < ch)70         ch = getchar();71     while ('0' <= ch && ch <= '9') {72         x = x * 10 + ch - '0';73         ch = getchar();74     }75     return x;76 }