The 11th Zhejiang Provincial Collegiate Programming Contest
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题意:输入n和m,接下来一个n*n的矩阵,a[i][j]表示第i道题放在第j个顺序做可以加a[i][j]的分数,问做完n道题所得分数大于等于m的概率。用分数表示,分母为上述满足题意的方案数,分子是总的方案数,输出最简形式。
思路:由于总的方案数为n! ,简化为求给一个n*n的矩阵,每一行每一列各选一个数使得n个数之和大于等于m的方案数。
n的范围是1 <= n <= 12,每一列选与不选分别用1和0表示,状态数最多达到1<<12。dp[sta][score]表示状态为i得分为j的方案数。当递推到任意一行i时,都有一个确定的状态数sta对应当前状态哪些列已经被选过。 在当前状态下,对于某一列j,若sta&(1<<j) == 0说明第j列还未选,继而可以由第j列来更新,否则说明第j列已经被选。
最后dp[ (1<<n) -1 ][ m ]表示每行每列各取一个数,最后取n个数并得分大于等于m的方案数。
- #include <stdio.h>
- #include <string.h>
- #include <algorithm>
- using namespace std;
- int dp[1<<12][510];
- int f[13];
- int a[13][13];
- int gcd(int a, int b)
- {
- if(b == 0) return a;
- return gcd(b,a%b);
- }
- int main()
- {
- int test;
- int n,m;
- f[0] = 1;
- for(int i = 1; i <= 12; i++)
- f[i] = f[i-1] * i;
- scanf("%d",&test);
- while(test--)
- {
- scanf("%d %d",&n,&m);
- for(int i = 1; i <= n; i++)
- for(int j = 1; j <= n; j++)
- scanf("%d",&a[i][j]);
- for(int i = 0; i < (1<<n); i++)
- for(int j = 0; j <= m; j++)
- dp[i][j] = 0;
- dp[0][0] = 1;
- for(int i = 0; i < (1<<n); i++)
- {
- int cnt = 0;
- for(int j = 1; j <= n; j++)
- {
- if(i & (1<<(j-1)) )
- cnt++;
- }
- for(int j = 1; j <= n; j++)
- {
- if(i & (1<<(j-1))) continue;
- for(int g = 0; g <= m; g++)
- {
- if(g + a[cnt+1][j] >= m)
- dp[i+(1<<(j-1))][m] += dp[i][g];
- else
- dp[i+(1<<(j-1))][g+a[cnt+1][j]] += dp[i][g];
- }
- }
- }
- if(dp[(1<<n)-1][m] == 0)
- printf("No solution\n");
- else
- {
- int tmp = gcd(f[n],dp[(1<<n)-1][m]);
- printf("%d/%d\n",f[n]/tmp, dp[(1<<n)-1][m]/tmp);
- }
- }
- return 0;
- }
Talented Chef
题意:给出n和m以及n个数,要求每次从n个数里取出m个数并减一,问最少需要几次把n个数减为0.
- #include <stdio.h>
- #include <string.h>
- #include <algorithm>
- #define eps 1e-8
- using namespace std;
- int main()
- {
- int test;
- scanf("%d",&test);
- while(test--)
- {
- int n,m;
- int maxnum = -1,sum = 0;
- scanf("%d %d",&n,&m);
- for(int i = 1; i <= n; i++)
- {
- int x;
- scanf("%d",&x);
- sum += x;
- maxnum = max(maxnum,x);
- }
- int ans = sum/m;
- if(sum%m != 0) ans++;
- if(ans < maxnum)
- printf("%d\n",maxnum);
- else printf("%d\n",ans);
- }
- return 0;
- }
Paint the Grid Reloaded(dfs+bfs)
题意:给出一个n*m的矩阵,只有‘O’和‘X’,每次可以选择一个四连块进行翻转,问使矩阵全部变为‘O’或‘X’需要的最少步数。
思路:dfs找出每个四连块并进行标号,缩点建图,每个点到其他所有点都有一个最大距离,找到这样一个点它到其他所有点的最大距离最短,该值就是最少需要翻转的次数。
用邻接矩阵建图一直TLE,改成前向星就A了。。。
- #include <stdio.h>
- #include <string.h>
- #include <algorithm>
- #include <queue>
- using namespace std;
- const int maxn = 1610;
- const int maxm = 42;
- const int INF = 0x3f3f3f3f;
- int dir[4][2] = {{-1,0},{1,0},{0,1},{0,-1}};
- int n,m;
- char s[maxm];
- int a[maxm][maxm],vis[maxm][maxn],scc;
- int p[maxn],cnt;
- int inque[maxn],dis[maxn];
- queue <int> que;
- struct node
- {
- int u,v,next;
- }edge[100000];
- void init()
- {
- memset(a,-1,sizeof(a));
- memset(vis,0,sizeof(vis));
- scc = 0;
- memset(p,-1,sizeof(p));
- cnt = 0;
- }
- int judge(int i,int j)
- {
- if(i >= 1 && i <= n && j >= 1 && j <= m)
- return 1;
- return 0;
- }
- void add(int u, int v)
- {
- edge[cnt] = (struct node){u,v,p[u]};
- p[u] = cnt++;
- }
- void dfs(int i, int j, int col)
- {
- if(vis[i][j] || a[i][j] != col)
- return;
- vis[i][j] = scc;
- dfs(i-1,j,col);
- dfs(i+1,j,col);
- dfs(i,j-1,col);
- dfs(i,j+1,col);
- }
- void build()
- {
- for(int i = 1; i <= n; i++)
- {
- for(int j = 1; j <= m; j++)
- {
- for(int k = 0; k < 4; k++)
- {
- int ii = i + dir[k][0];
- int jj = j + dir[k][1];
- if(judge(ii,jj) && vis[ii][jj] != vis[i][j])
- add(vis[i][j],vis[ii][jj]);
- }
- }
- }
- }
- int bfs(int s)
- {
- while(!que.empty()) que.pop();
- memset(inque,0,sizeof(inque));
- memset(dis,INF,sizeof(dis));
- int ans = 0;
- dis[s] = 0;
- inque[s] = 1;
- que.push(s);
- while(!que.empty())
- {
- int u = que.front();
- que.pop();
- for(int i = p[u]; i != -1; i = edge[i].next)
- {
- int v = edge[i].v;
- if(!inque[v])
- {
- inque[v] = 1;
- que.push(v);
- dis[v] = min(dis[v],dis[u]+1);
- ans = max(ans,dis[v]);
- }
- }
- }
- return ans;
- }
- int main()
- {
- int test;
- scanf("%d",&test);
- while(test--)
- {
- init();
- scanf("%d %d",&n,&m);
- for(int i = 1; i <= n; i++)
- {
- scanf("%s",s+1);
- for(int j = 1; j <= m; j++)
- if(s[j] == 'O')
- a[i][j] = 0;
- else a[i][j] = 1;
- }
- for(int i = 1; i <= n; i++)
- {
- for(int j = 1; j <= m; j++)
- {
- if(!vis[i][j])
- {
- scc++;
- dfs(i,j,a[i][j]);
- }
- }
- }
- build();
- int ans = INF;
- for(int i = 1; i <= scc; i++)
- {
- int res = bfs(i);
- ans = min(ans,res);
- }
- printf("%d\n",ans);
- }
- return 0;
- }
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