SGU125 Shtirlits

题目大意

有一个N*N的矩阵，每个格子里有一个值
已知对于每个格子，四周有多少个格子的值大于它
构造这个矩阵

算法思路

DFS+剪枝，按照从上到下、从左到右的顺序，枚举每个格子的值后有如下剪枝

• 判断上方和左边两个格子中大于它的个数，超过则非法，加上右边和下面的不够也非法
• 对于非第一行，判断上方的格子是否符合，不符合则非法
• 对于最后一行，判断左边的格子是否符合，不符合则非法

时间复杂度: 稍低于O(10N×N)

代码

/** * Copyright © 2015 Authors. All rights reserved. *  * FileName: 125.cpp * Author: Beiyu Li <[e-mail]> * Date: 2015-06-04 */#include <bits/stdc++.h>using namespace std;#define rep(i,n) for (int i = 0; i < (n); ++i)#define For(i,s,t) for (int i = (s); i <= (t); ++i)#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)typedef long long LL;typedef pair<int, int> Pii;const int inf = 0x3f3f3f3f;const LL infLL = 0x3f3f3f3f3f3f3f3fLL;const int maxn = 3 + 5;int n;int a[maxn][maxn], b[maxn][maxn];int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};bool inside(int x, int y) { return 0 <= x && x < n && 0 <= y && y < n; }int calc(int x, int y){        int c = 0;        rep(i,4) {                int nx = x + dx[i], ny = y + dy[i];                if (inside(nx, ny) && a[nx][ny] > a[x][y]) ++c;        }        return c;}bool dfs(int i, int j){        if (i == n && calc(n - 1, n - 1) == b[n-1][n-1]) return true;        for (a[i][j] = 0; a[i][j] < 10; ++a[i][j]) {                int c = 0;                if (i && a[i-1][j] > a[i][j]) ++c;                if (j && a[i][j-1] > a[i][j]) ++c;                if (c > b[i][j]) continue;                if (c + (i < n - 1? 2: 1) < b[i][j]) return false;                if (i) {                        c = calc(i - 1, j);                        if (c < b[i-1][j]) continue;                        if (c > b[i-1][j]) return false;                }                if (i == n - 1 && j) {                        c = calc(i, j - 1);                        if (c < b[i][j-1]) continue;                        if (c > b[i][j-1]) return false;                }                if (dfs(i + (j + 1) / n, (j + 1) % n)) return true;        }        return false;}int main(){        scanf("%d", &n);        rep(i,n) rep(j,n) scanf("%d", &b[i][j]);        bool ok = true;        rep(i,n) rep(j,n) {                a[i][j] = -1;                if (calc(i, j) < b[i][j]) ok = false;                a[i][j] = 0;        }        if (ok && dfs(0, 0)) {                rep(i,n) rep(j,n) printf("%d%c", a[i][j], " \n"[j==n-1]);        } else {                puts("NO SOLUTION");        }        return 0;}

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