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发几个以前写的拓扑排序，回顾一下。
拓扑排序，一般不会单独考，主要要求还是掌握好这个概念，有个感性的认识，以及能快速的写出求拓扑排序的程序，进而继续接下来对图的处理，或是比如dp之类的算法，又或者是判断有无环之类。求拓扑序主要就是运用队列，push入度为0的点，删掉它们出去的边，重复这个操作。像要是求字典序最小，就可以用优先队列。
TOJ 3993 求字典序最小的拓扑排序
1 #include<cstdio> 2 #include<queue> 3 #include<cstring> 4 using namespace std; 5 6 int ans[110], in[110]; 7 bool e[110][110]; 8 int n, m; 9 int main() 10 { 11 int T, x, y; 12 scanf("%d", &T); 13 while(T--) 14 { 15 memset(e, 0, sizeof(e)); 16 memset(in, 0, sizeof(in)); 17 scanf("%d %d", &n, &m); 18 for (int i = 0; i < m; i++){ 19 scanf("%d %d", &x, &y); 20 if (!e[x][y]){ 21 e[x][y] = 1; 22 in[y]++; 23 } 24 } 25 priority_queue<int, vector<int>, greater<int> > q; 26 for (int i = 1; i <= n; i++) 27 if (!in[i]) q.push(i); 28 int cnt = 0; 29 while(!q.empty()){ 30 int u = q.top(); 31 q.pop(); 32 ans[++cnt] = u; 33 for (int i = 1; i <= n; i++) 34 if (e[u][i]){ 35 in[i]--; 36 if (!in[i]) q.push(i); 37 } 38 } 39 if (cnt == n){ 40 for (int i = 1; i <= n; i++) 41 printf("%d ", ans[i]); 42 printf("\n"); 43 } 44 else printf("Low IQ\n"); 45 } 46 return 0; 47 }
POJ 3687 反向建图求拓扑排序
1 #include<cstring> 2 #include<cstdio> 3 #include<queue> 4 using namespace std; 5 6 int ans[210], in[210]; 7 bool e[210][210]; 8 int main() 9 { 10 int T, n, m, x, y; 11 scanf("%d", &T); 12 while(T--) 13 { 14 memset(e, false, sizeof(e)); 15 memset(in, 0, sizeof(in)); 16 scanf("%d %d", &n, &m); 17 for (int i = 0; i < m; i++){ 18 scanf("%d %d", &x, &y); 19 if (!e[y][x]){ 20 e[y][x] = true; 21 in[x] ++; 22 } 23 } 24 priority_queue<int> q; 25 for (int i = 1; i <= n; i++) 26 if (!in[i]) q.push(i); 27 int cnt = n; 28 while(!q.empty()){ 29 int u = q.top(); q.pop(); 30 ans[u] = cnt --; 31 for (int i = 1; i <= n; i++) 32 if (e[u][i]){ 33 in[i] --; 34 if (!in[i]) q.push(i); 35 } 36 } 37 if (cnt == 0){ 38 for (int i = 1; i < n; i++) 39 printf("%d ", ans[i]); 40 printf("%d\n", ans[n]); 41 } 42 else printf("-1\n"); 43 } 44 return 0; 45 }
POJ 1270 按字典序输出所有拓扑排序，这个代码写得比较dirty
1 #include<cstdio> 2 #include<cstring> 3 #include<queue> 4 #include<algorithm> 5 #include<string> 6 #include<iostream> 7 using namespace std; 8 9 char s1[1000], s2[1000]; 10 string ans[400]; 11 int n, cnt; 12 bool vis[50], e[50][50]; 13 int in[50], a[50]; 14 void dfs(int dep, string ss) 15 { 16 if (dep == n){ 17 ans[cnt++] = ss; 18 return; 19 } 20 for (int i = 0; i < n; i++) 21 if (!in[a[i]] && !vis[a[i]]){ 22 vis[a[i]] = true; 23 char tmp = a[i] + ‘a‘ - 1; 24 ss.append(1, tmp); 25 for (int j = 0; j < n; j++) 26 if (e[a[i]][a[j]]) in[a[j]] --; 27 28 dfs(dep+1, ss); 29 30 vis[a[i]] = false; 31 ss.erase(ss.length()-1, 1); 32 for (int j = 0; j < n; j++) 33 if (e[a[i]][a[j]]) in[a[j]] ++; 34 } 35 return; 36 } 37 bool cmp(string a, string b){ 38 return a < b; 39 } 40 int main() 41 { 42 bool flag = true; 43 while(cin.getline(s1, 1000)) 44 { 45 if (flag) flag = false; 46 else printf("\n"); 47 cin.getline(s2, 1000); 48 n = 0; 49 for (int i = 0; i < strlen(s1); i++){ 50 if (s1[i] != ‘ ‘){ 51 a[n++] = s1[i] - ‘a‘ + 1; 52 } 53 } 54 // for (int i = 0; i < n; i++) 55 // printf("%d\n", a[i]); 56 memset(e, 0, sizeof(e)); 57 memset(in, 0, sizeof(in)); 58 memset(vis, false, sizeof(vis)); 59 int p1 = 0, p2 = 0; 60 for (int i = 0; i < strlen(s2); i++){ 61 if (s2[i] != ‘ ‘){ 62 if (p1 == 0){ 63 p1 = s2[i] - ‘a‘ + 1; 64 } 65 else{ 66 p2 = s2[i] - ‘a‘ + 1; 67 e[p1][p2] = true; 68 in[p2] ++; 69 p1 = p2 = 0; 70 } 71 } 72 } 73 string ss = ""; 74 cnt = 0; 75 dfs(0, ss); 76 sort(ans, ans+cnt, cmp); 77 for (int i = 0; i < cnt; i++) 78 cout << ans[i] << endl; 79 } 80 return 0; 81 }
HDU 1285 求字典序最小拓扑序
1 #include<cstring> 2 #include<cstdio> 3 #include<queue> 4 using namespace std; 5 6 int en[510], in[510]; 7 int n, m, x, y, esize; 8 struct edge{ 9 int n, v; 10 } e[30000]; 11 void insert(int u, int v) 12 { 13 esize ++; 14 e[esize].v = v; 15 e[esize].n = en[u]; 16 en[u] = esize; 17 } 18 int main() 19 { 20 while(scanf("%d %d", &n, &m) != EOF) 21 { 22 memset(en, -1, sizeof(en)); 23 memset(in, 0, sizeof(in)); 24 esize = -1; 25 for(int i = 0; i < m; i++){ 26 scanf("%d %d", &x, &y); 27 insert(x, y); 28 in[y]++; 29 } 30 priority_queue<int, vector<int>, greater<int> > q; 31 for (int i = 1; i <= n; i++){ 32 if (!in[i]) q.push(i); 33 } 34 bool flag = true; 35 while(!q.empty()){ 36 int t = q.top(); q.pop(); 37 if (flag){ 38 flag = false; 39 printf("%d", t); 40 } 41 else printf(" %d", t); 42 for (int tt = en[t]; tt != -1; tt = e[tt].n){ 43 int v = e[tt].v; 44 in[v]--; 45 if (!in[v]) q.push(v); 46 } 47 } 48 printf("\n"); 49 } 50 return 0; 51 }
HDU 3231 在三个坐标轴方向上做拓扑排序，对这题还挺有印象，当时wa了好几次卡了很久，记得这题还是某个区预赛的题。不知道真到比赛，会是什么样。我能很快的做出来吗？
1 #include<cstdio> 2 #include<cstring> 3 #include<queue> 4 using namespace std; 5 6 struct edge{ 7 int n, v; 8 } e[3][603000]; 9 int ans[3000][3], en[3][3000], in[3][3000]; 10 int n, r, esize[3]; 11 void insert(int t, int u, int v) 12 { 13 e[t][esize[t]].v = v; 14 e[t][esize[t]].n = en[t][u]; 15 en[t][u] = esize[t]; 16 esize[t] ++; 17 in[t][v]++; 18 } 19 int main() 20 { 21 int x, y, ca = 0; 22 char ch[10]; 23 while(scanf("%d %d", &n, &r) && (n + r)) 24 { 25 ca ++; 26 memset(in, 0, sizeof(in)); 27 memset(en, -1, sizeof(en)); 28 memset(esize, 0, sizeof(esize)); 29 for (int i = 1; i <= n; i++){ 30 insert(0, i, i+n); 31 insert(1, i, i+n); 32 insert(2, i, i+n); 33 } 34 for (int i = 0; i < r; i++){ 35 //getchar(); 36 scanf("%s %d %d", ch, &x, &y); 37 if (ch[0] == ‘I‘){ 38 insert(0, x, y+n); 39 insert(1, x, y+n); 40 insert(2, x, y+n); 41 insert(0, y, x+n); 42 insert(1, y, x+n); 43 insert(2, y, x+n); 44 } 45 else if (ch[0] == ‘X‘){ 46 insert(0, x+n, y); 47 } 48 else if (ch[0] == ‘Y‘){ 49 insert(1, x+n, y); 50 } 51 else if (ch[0] == ‘Z‘) 52 insert(2, x+n, y); 53 } 54 bool findans = true; 55 for (int t = 0; t < 3; t++){ 56 queue<int> q; 57 for (int i = 1; i <= n * 2; i++){ 58 if (in[t][i] == 0) {//printf("t:%d %d\n", t, i); 59 q.push(i); 60 } 61 } 62 int u, v, p; 63 int cnt = 0; 64 while(!q.empty()){ 65 int u = q.front(); q.pop(); 66 ans[u][t] = ++cnt; 67 for (p = en[t][u]; p != -1; p = e[t][p].n){ 68 v = e[t][p].v; 69 in[t][v] --; 70 if (!in[t][v]){//printf("t:%d %d\n", t, v); 71 q.push(v); 72 } 73 } 74 } 75 //printf("%d\n", cnt); 76 if (cnt != (n*2)) {findans = false; break;} 77 } 78 if (findans){ 79 printf("Case %d: POSSIBLE\n", ca); 80 for (int i = 1; i <= n; i++) 81 printf("%d %d %d %d %d %d\n", ans[i][0], ans[i][1], ans[i][2], 82 ans[i+n][0], ans[i+n][1], ans[i+n][2]); 83 } 84 else 85 printf("Case %d: IMPOSSIBLE\n", ca); 86 if (ca >= 1) printf("\n"); 87 } 88 return 0; 89 }
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标签：style blog http color os io for art
原文地址：http://www.cnblogs.com/james47/p/3903461.html
