The 11th Zhejiang Provincial Collegiate Programming Contest
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Problem Arrangement(状态压缩+递推)
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3777
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
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3778
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-8using 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)
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3781
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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