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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时间： 2024-10-11 02:48:53