题意：给一个的格子图，有 n 行单元格，每行有a[i]个格子，要求往格子中填1~m的数字，要求每个数字大于等于左边的数字，大于上边的数字，问有多少种填充方法。
析：感觉像个DP，但是不会啊。。。就想暴力试试，反正数据量看起来不大才7，但是。。。TLE了，又换了一个暴力方法，2秒多过了，差点啊。
其实这是一个状压DP，dp[i][s]表示在第 i 列，在集合 s 中有方法数，那么怎么转移呢，这个还是挺简单的，就是判断第i+1列是不是比第 i 列都大于等于就ok了，
输入时先把行，转化成列，再计算，初始化就是第一列喽，假设什么组合都行。
代码如下：
暴力代码：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 450 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int w[10], h[10];
int ans = 0, cnt[10][10];
void dfs(int r, int c, int mx){
if(c > m){
++ans; return ;
}
int mm = Min(n, n-h[c]+r);
for(int i = Max(mx+1, cnt[r][c-1]); i <= mm; ++i){
cnt[r][c] = i;
if(r == h[c]) dfs(1, c+1, 0);
else dfs(r+1, c, i);
}
}
int main(){
int k;
while(scanf("%d", &k) == 1){
memset(h, 0, sizeof h);
m = 0;
for(int i = 1; i <= k; ++i){
int x;
scanf("%d", &x);
m = Max(m, x);
for(int j = 1; j <= x; ++j)
++h[j];
}
scanf("%d", &n);
ans = 0;
memset(cnt, 0, sizeof cnt);
dfs(1, 1, 0);
printf("%d
", ans);
}
return 0;
}
状压DP代码:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 8;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[10];
int dp[maxn][1<<maxn];
inline int lowbit(int x){
return x & (-x);
}
inline int bitcount(int i){//返回 i 这个数二进制数中有几个1
int ans = 0;
while(i) ++ans, i -= lowbit(i);
return ans;
}
inline bool judge(int j, int k){//判断第 j 列是不是小于等于第 k 列
int tj[10], tk[10];
int cntj = 0, cntk = 0;
for(int i = 0; i < m; ++i){
if(j>>i&1) tj[cntj++] = i;
if(k>>i&1) tk[cntk++] = i;
}
for(int i = 0; i < cntk; ++i)
if(tj[i] > tk[i]) return false;
return true;
}
int main(){
while(scanf("%d", &n) == 1){
memset(a, 0, sizeof a);
int t = 0, x;
for(int i = 0; i < n; ++i){
scanf("%d", &x);
t = Max(t, x);
for(int j = 1; j <= x; ++j)
++a[j];
}
memset(dp, 0, sizeof dp);
scanf("%d", &m);
int len = 1<<m;
for(int i = 0; i < len; ++i)//初始化第 1 列
if(bitcount(i) == a[1]) dp[1][i] = 1;
for(int i = 1; i < t; ++i){
for(int j = 0; j < len; ++j){
if(bitcount(j) != a[i] || !dp[i][j]) continue;
for(int k = 0; k < len; ++k)
if(bitcount(k) == a[i+1] && judge(j, k)) dp[i+1][k] += dp[i][j];
}
}
int ans = 0;
for(int i = 0; i < len; ++i) ans += dp[t][i];
printf("%d
", ans);
}
return 0;
}