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这题开始的时候想不到怎么调整每个grid的实际淹没时间，于是只好找了下watashi的题解，发现这个操作还是挺简单的。
debug这题累死了。。解释还是移步看watashi大神的吧。。
除了开始的时候这个调整，后面并查集的部分是相当容易想到的，其实就是用并查集了统计每一个块的个数。
代码如下：
1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 8 using namespace std; 9 10 int n, m; 11 inline int con(int x, int y) { return x * m + y;} 12 inline bool inmat(int x, int y) { return 0 <= x && x < n && 0 <= y && y < m;} 13 14 const int dx[4] = { 0, -1, 0, 1}; 15 const int dy[4] = { -1, 0, 1, 0}; 16 const int N = 555; 17 const int M = N * N; 18 19 struct MFS { 20 int fa[M], cnt[M]; 21 int fx, fy; 22 void init() { for (int i = 0; i < M; i++) fa[i] = i, cnt[i] = 1;} 23 int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]);} 24 void merge(int x, int y) { 25 fx = find(x), fy = find(y); 26 if (fx == fy) return ; 27 fa[fy] = fx; 28 cnt[fx] += cnt[fy]; 29 } 30 bool same(int x, int y) { return find(x) == find(y);} 31 int count(int x) { return cnt[find(x)];} 32 } mfs; 33 bool vis[N][N]; 34 int qry[11111][2], hmk[11111]; 35 36 struct Grid { 37 int h, x, y; 38 Grid() {} 39 Grid(int h, int x, int y) : h(h), x(x), y(y) {} 40 bool operator < (Grid a) const { return h > a.h;} 41 } grid[M]; 42 43 int mat[N][N]; 44 map id; 45 map cur, rec[11111]; 46 priority_queue pq; 47 48 int bitcnt(int x) { 49 int ret = 0; 50 for (int i = 0; i < 32; i++) if (x & 1 << i) ret++; 51 return ret; 52 } 53 54 int main() { 55 // freopen("in", "r", stdin); 56 // freopen("out", "w", stdout); 57 int matsz, t; 58 int cx, cy, nx, ny; 59 while (~scanf("%d%d", &n, &m)) { 60 matsz = n * m, t = 0; 61 while (!pq.empty()) pq.pop(); 62 memset(vis, 0, sizeof(vis)); 63 for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) { 64 scanf("%d", &mat[i][j]); 65 if (i == 0 || j == 0 || i == n - 1 || j == m - 1) { 66 grid[t] = Grid(mat[i][j], i, j); 67 pq.push(grid[t]); 68 t++; 69 vis[i][j] = true; 70 } 71 } 72 while (!pq.empty()) { 73 cx = pq.top().x, cy = pq.top().y; 74 pq.pop(); 75 for (int i = 0; i < 4; i++) { 76 nx = cx + dx[i]; 77 ny = cy + dy[i]; 78 if (inmat(nx, ny)) { 79 if (vis[nx][ny]) continue; 80 mat[nx][ny] = max(mat[nx][ny], mat[cx][cy]); 81 grid[t] = Grid(mat[nx][ny], nx, ny); 82 pq.push(grid[t]); 83 t++; 84 vis[nx][ny] = true; 85 } 86 } 87 } 88 sort(grid, grid + matsz); 89 int k; 90 scanf("%d", &k); 91 for (int i = 0; i < k; i++) { 92 for (int j = 0; j < 2; j++) scanf("%d", &qry[i][j]); 93 hmk[i] = qry[i][0]; 94 } 95 sort(hmk, hmk + k); 96 int sz = unique(hmk, hmk + k) - hmk; 97 reverse(hmk, hmk + sz); 98 id.clear(); 99 cur.clear();100 memset(vis, 0, sizeof(vis));101 mfs.init();102 for (int i = 0, c = 0, a, b; i < sz; i++) {103 id[hmk[i]] = i;104 while (c < matsz && grid[c].h > hmk[i]) {105 cx = grid[c].x, cy = grid[c].y;106 vis[cx][cy] = true;107 cur[1]++;108 for (int d = 0; d < 4; d++) {109 nx = cx + dx[d], ny = cy + dy[d];110 if (!inmat(nx, ny) || !vis[nx][ny]) continue;111 a = con(cx, cy), b = con(nx, ny);112 if (!mfs.same(a, b)) {113 int ca = mfs.count(a);114 int cb = mfs.count(b);115 cur[ca]--;116 if (cur[ca] == 0) cur.erase(ca);117 cur[cb]--;118 if (cur[cb] == 0) cur.erase(cb);119 mfs.merge(a, b);120 cur[mfs.count(a)]++;121 }122 }123 c++;124 }125 rec[i] = cur;126 }127 map ::iterator mii;128 int ans, tid;129 for (int i = 0; i < k; i++) {130 ans = 0, tid = id[qry[i][0]];131 for (mii = rec[tid].begin(); mii != rec[tid].end(); mii++) {132 ans += (*mii).second << bitcnt((*mii).first & qry[i][1]);133 }134 printf("%d\n", ans);135 }136 }137 return 0;138 }
转载于:https://www.cnblogs.com/LyonLys/p/zoj_3354_Lyon.html
