Hdu OJ 5113 Black And White (2014ACM/ICPC亚洲区北京站) （搜索）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5113

题目大意：有k种颜色的方块，每种颜色有a_i个， 现在有n*m的矩阵， 问这k种颜色的方块能否使任意两个相连的方块颜色不一样填满整个矩阵，如果可以输出任意一种。

解题思路：由于n，m在[1, 5]内， 所以直接暴力枚举； 需要注意的是：需要剪枝一下。

代码如下：

复制代码#include<stdio.h> #include<string.h> #include<cmath> #include<queue> #include<algorithm> using namespace std; typedef long long ll; const int N = 103; int n, m, k; int a[N], mp[N][N]; bool is; bool judge(int x, int y, int s) { bool p = true, q = true; if(x - 1 >= 1 && mp[x-1][y] == s) p = false; if(y - 1 >= 1 && mp[x][y-1] == s) q = false; return p && q; } void dfs(int x, int y, int cou) { int sum = 0, mx = 0; for(int i=1; i<=k; ++ i) { sum += a[i]; if(a[i] > mx) mx = a[i]; } if(sum - mx < mx - 1) return; if(cou == n*m) { is = true; return ; } for(int i=1; i<=k; ++ i) { if(a[i] && judge(x, y, i)) { a[i] --; mp[x][y] = i; if(y == m) dfs(x+1, 1, cou + 1); else dfs(x, y+1, cou + 1); if(is) return ; a[i] ++; } } } void solve(int cases) { scanf("%d%d%d", &n, &m, &k); int sum = 0, mx = 0; for(int i=1; i<=k; ++ i) { scanf("%d", &a[i]); sum += a[i]; if(a[i] > mx) mx = a[i]; } printf("Case #%d:n", cases); if(sum - mx < mx - 1) printf("NOn"); else { printf("YESn"); memset(mp, -1, sizeof(mp)); is = false; dfs(1, 1, 0); for(int i=1; i<=n; ++ i) for(int j=1; j<=m; ++ j) printf(j == m? "%dn" : "%d ", mp[i][j]); } } int main() { int t; scanf("%d", &t); for(int i = 1; i <= t; ++ i) { solve(i); } return 0; }

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#include<stdio.h>
#include<string.h>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 103;
int n, m, k;
int a[N], mp[N][N];
bool is;
bool judge(int x, int y, int s)
{
bool p = true, q = true;
if(x - 1 >= 1 && mp[x-1][y] == s)
p = false;
if(y - 1 >= 1 && mp[x][y-1] == s)
q = false;
return p && q;
}
void dfs(int x, int y, int cou)
{
int sum = 0, mx = 0;
for(int i=1; i<=k; ++ i)
{
sum += a[i];
if(a[i] > mx)
mx = a[i];
}
if(sum - mx < mx - 1)
return;
if(cou == n*m)
{
is = true;
return ;
}
for(int i=1; i<=k; ++ i)
{
if(a[i] && judge(x, y, i))
{
a[i] --;
mp[x][y] = i;
if(y == m)
dfs(x+1, 1, cou + 1);
else
dfs(x, y+1, cou + 1);
if(is)
return ;
a[i] ++;
}
}
}
void solve(int cases)
{
scanf("%d%d%d", &n, &m, &k);
int sum = 0, mx = 0;
for(int i=1; i<=k; ++ i)
{
scanf("%d", &a[i]);
sum += a[i];
if(a[i] > mx)
mx = a[i];
}
printf("Case #%d:n", cases);
if(sum - mx < mx - 1)
printf("NOn");
else
{
printf("YESn");
memset(mp, -1, sizeof(mp));
is = false;
dfs(1, 1, 0);
for(int i=1; i<=n; ++ i)
for(int j=1; j<=m; ++ j)
printf(j == m? "%dn" : "%d ", mp[i][j]);
}
}
int main()
{
int t;
scanf("%d", &t);
for(int i = 1; i <= t; ++ i)
{
solve(i);
}
return 0;
}