有很多奇妙的题.....
题目VJ上做了些.
近期在POJ和HDU上做一些题,参考 http://blog.csdn.net/shahdza/article/details/7779389
HDU 1285 裸题不讲=w=很久以前用Vector建图的时候就A了......
HDU 2647 就是判断一下是否有拓扑序列(这个可以用强连通分量Tarjan做=w=但是拓扑排写起来比较简单...)
然后总代价就是按照排序层数(具体看程序,我们把能够同时存在于队列中的元素表上同一个dep)算....
1 #include <cstdio>
2 #include <fstream>
3 #include <iostream>
4
5 #include <cstdlib>
6 #include <cstring>
7 #include <algorithm>
8 #include <cmath>
9
10 #include <queue>
11 #include <vector>
12 #include <map>
13 #include <set>
14 #include <stack>
15 #include <list>
16
17 typedef unsigned int uint;
18 typedef long long int ll;
19 typedef unsigned long long int ull;
20 typedef double db;
21
22 using namespace std;
23
24 inline int getint()
25 {
26 int res=0;
27 char c=getchar();
28 bool mi=false;
29 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
30 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
31 return mi ? -res : res;
32 }
33 inline ll getll()
34 {
35 ll res=0;
36 char c=getchar();
37 bool mi=false;
38 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
39 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
40 return mi ? -res : res;
41 }
42
43 //==============================================================================
44 //==============================================================================
45 //==============================================================================
46 //==============================================================================
47
48
49 struct edge
50 {
51 int in;
52 edge*nxt;
53 }pool[20050];
54 edge*et;
55 edge*eds[10050];
56 void addedge(int a,int b)
57 { et->in=b; et->nxt=eds[a]; eds[a]=et++; }
58 #define FOREACH_EDGE(i,j) for(edge*i=eds[j];i;i=i->nxt)
59
60
61 int n,m;
62
63 int dep[10050];
64 int deg[10050];
65 bool used[10050];
66
67
68 int main()
69 {
70 while(scanf("%d%d",&n,&m)>0)
71 {
72 et=pool;
73 memset(eds,0,sizeof(edge*)*(n+1));
74 memset(dep,0,sizeof(int)*(n+1));
75 memset(deg,0,sizeof(int)*(n+1));
76 memset(used,0,sizeof(bool)*(n+1));
77
78 for(int i=0;i<m;i++)
79 {
80 int a=getint()-1;
81 int b=getint()-1;
82 addedge(b,a);
83 deg[a]++;
84 }
85
86 queue<int> q;
87 for(int i=0;i<n;i++)
88 if(deg[i]==0) q.push(i);
89
90 while(!q.empty())
91 {
92 int x=q.front(); q.pop();
93 used[x]=true;
94 FOREACH_EDGE(e,x)
95 {
96 deg[e->in]--;
97 if(deg[e->in]==0)
98 {
99 dep[e->in]=dep[x]+1;
100 q.push(e->in);
101 }
102 }
103 }
104
105 bool ok=true;
106 for(int i=0;i<n;i++)
107 if(!used[i]) { ok=false; break; }
108
109 if(!ok)
110 {
111 printf("-1
");
112 continue;
113 }
114
115 int res=0;
116 for(int i=0;i<n;i++)
117 res+=dep[i]+888;
118
119 printf("%d
",res);
120 }
121
122 return 0;
123 }
HDU 1811
判断是否存在拓扑序,以及是否存在确定的拓扑序.
判断是否存在,就是按照上面的,能把所有点压入队列就肯定存在可行拓扑序列.
(一般有环就说明限制条件存在矛盾,这个环中的任意一个节点都不能被压入队列)
拓扑序列只有一个的条件是每时每刻队列中仅有一个节点.(否则这两个节点的顺序不定.)
处理那个等号非常麻烦.三次提交都错在那里. 这个题看起来还可以用差分约束.
1 #include <cstdio>
2 #include <fstream>
3 #include <iostream>
4
5 #include <cstdlib>
6 #include <cstring>
7 #include <algorithm>
8 #include <cmath>
9
10 #include <queue>
11 #include <vector>
12 #include <map>
13 #include <set>
14 #include <stack>
15 #include <list>
16
17 typedef unsigned int uint;
18 typedef long long int ll;
19 typedef unsigned long long int ull;
20 typedef double db;
21
22 using namespace std;
23
24 inline int getint()
25 {
26 int res=0;
27 char c=getchar();
28 bool mi=false;
29 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
30 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
31 return mi ? -res : res;
32 }
33 inline ll getll()
34 {
35 ll res=0;
36 char c=getchar();
37 bool mi=false;
38 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
39 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
40 return mi ? -res : res;
41 }
42
43 //==============================================================================
44 //==============================================================================
45 //==============================================================================
46 //==============================================================================
47
48 struct edge
49 {
50 int in;
51 edge*nxt;
52 }pool[20050];
53 edge*et;
54 edge*eds[2][10050];
55 void addedge(int i,int j,int k)
56 { et->in=j; et->nxt=eds[k][i]; eds[k][i]=et++; }
57 #define FOREACH_EDGE(i,j,f) for(edge*i=eds[f][j];i;i=i->nxt)
58
59 //union find
60 int f[10050];
61 void INIT(int size)
62 { for(int i=0;i<size;i++) f[i]=i; }
63 int findf(int x)
64 { return f[x]==x ? x : f[x]=findf(f[x]); }
65 void connect(int a,int b)
66 { f[findf(a)]=findf(b); }
67
68 int block[10050];
69 int btot;
70 bool used[10050];
71 int deg[10050];
72
73 int n,m;
74
75 int A[20050];
76 int B[20050];
77 char C[20050];
78
79 int main()
80 {
81 while(scanf("%d%d",&n,&m)>0)
82 {
83 INIT(n+1);
84 memset(eds[1],0,sizeof(edge*)*(n+1));
85 memset(block,0xFF,sizeof(int)*(n+1));
86 et=pool;
87 btot=0;
88
89 bool unc=false;
90 bool conf=false;
91
92 for(int i=0;i<m;i++)
93 {
94 scanf("%d %c %d",&A[i],&C[i],&B[i]);
95 if(C[i]=='=') connect(A[i],B[i]);
96 }
97
98 for(int i=0;i<m;i++)
99 if(C[i]!='=')
100 {
101 if(findf(A[i])==findf(B[i])) conf=true;
102
103 if(C[i]=='>') addedge(A[i],B[i],1);
104 else if(C[i]=='<') addedge(B[i],A[i],1);
105 }
106
107 if(conf) { printf("CONFLICT
"); continue; }
108
109 for(int i=0;i<n;i++)
110 {
111 if(block[findf(i)]==-1)
112 block[findf(i)]=btot++;
113 block[i]=block[findf(i)];
114 }
115
116 memset(eds[0],0,sizeof(edge*)*(btot+1));
117 memset(deg,0,sizeof(int)*(btot+1));
118 memset(used,0,sizeof(bool)*(btot+1));
119
120 for(int i=0;i<n;i++)
121 FOREACH_EDGE(e,i,1)
122 if(block[i]!=block[e->in])
123 {
124 addedge(block[i],block[e->in],0);
125 deg[block[e->in]]++;
126 }
127
128 queue<int> q;
129 for(int i=0;i<btot;i++)
130 if(deg[i]==0) q.push(i);
131
132 while(!q.empty())
133 {
134 if(q.size()>1) unc=true;
135 int x=q.front(); q.pop();
136 used[x]=true;
137 FOREACH_EDGE(e,x,0)
138 {
139 deg[e->in]--;
140 if(deg[e->in]==0)
141 q.push(e->in);
142 }
143 }
144
145 for(int i=0;i<btot;i++)
146 if(!used[i]) { conf=true; break; }
147
148 if(conf) printf("CONFLICT
");
149 else if(unc) printf("UNCERTAIN
");
150 else printf("OK
");
151
152 }
153
154 return 0;
155 }
POJ 3287
裸的拓扑标号. 要求优先级.
做的时候,把边反向,做拓扑排序,然后逆序标号.
千万想清楚程序在干什么! 这道题是 "对标号进行限制" 以及 "对每个标号进行赋权" 然后 "输出标号的权值".
1 #include <cstdio>
2 #include <fstream>
3 #include <iostream>
4
5 #include <cstdlib>
6 #include <cstring>
7 #include <algorithm>
8 #include <cmath>
9
10 #include <queue>
11 #include <vector>
12 #include <map>
13 #include <set>
14 #include <stack>
15 #include <list>
16
17 typedef unsigned int uint;
18 typedef long long int ll;
19 typedef unsigned long long int ull;
20 typedef double db;
21
22 using namespace std;
23
24 inline int getint()
25 {
26 int res=0;
27 char c=getchar();
28 bool mi=false;
29 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
30 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
31 return mi ? -res : res;
32 }
33 inline ll getll()
34 {
35 ll res=0;
36 char c=getchar();
37 bool mi=false;
38 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
39 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
40 return mi ? -res : res;
41 }
42
43 db eps=1e-20;
44 inline bool feq(db a,db b)
45 { return fabs(a-b)<eps; }
46
47 template<typename Type>
48 inline Type avg(const Type a,const Type b)
49 { return a+((b-a)/2); }
50
51 //==========================================================
52 //==========================================================
53 //==========================================================
54 //==========================================================
55
56
57 struct edge
58 {
59 int in;
60 edge*nxt;
61 }pool[40050];
62 edge*et=pool;
63 edge*eds[205];
64 void addedge(int a,int b)
65 { et->in=b; et->nxt=eds[a]; eds[a]=et++; }
66 #define FOREACH_EDGE(i,j) for(edge*i=eds[j];i;i=i->nxt)
67
68
69 int n,m;
70 int deg[205];
71 int label[205];
72
73 int main()
74 {
75 int T=getint();
76 while(T--)
77 {
78 n=getint();
79 m=getint();
80
81 memset(eds,0,sizeof(edge*)*(n+1));
82 memset(deg,0,sizeof(int)*(n+1));
83 memset(label,0,sizeof(int)*(n+1));
84 et=pool;
85
86 int tot=0;
87
88 for(int i=0;i<m;i++)
89 {
90 int a=getint()-1;
91 int b=getint()-1;
92 addedge(b,a);
93 deg[a]++;
94 }
95
96 priority_queue<int> q;
97 for(int i=0;i<n;i++)
98 if(deg[i]==0) q.push(i);
99
100 while(!q.empty())
101 {
102 int x=q.top(); q.pop();
103 label[x]=n-(tot++);
104 FOREACH_EDGE(e,x)
105 {
106 deg[e->in]--;
107 if(deg[e->in]==0)
108 q.push(e->in);
109 }
110 }
111
112 if(tot!=n) printf("-1
");
113 else
114 {
115 printf("%d",label[0]);
116 for(int i=1;i<n;i++)
117 printf(" %d",label[i]);
118 printf("
");
119 }
120 }
121
122 return 0;
123 }
POJ 1270
枚举拓扑序并输出.
首先读入比较坑.....WA掉一次的原因是没排序...题目要求不是按照读入的字典序而是字母字典序.....果断写了个冒泡上去......然后就A了.....
这个算是拓扑排序么?唉随便吧....
1 #include <cstdio>
2 #include <fstream>
3 #include <iostream>
4
5 #include <cstdlib>
6 #include <cstring>
7 #include <algorithm>
8 #include <cmath>
9
10 #include <queue>
11 #include <vector>
12 #include <map>
13 #include <set>
14 #include <stack>
15 #include <list>
16
17 typedef unsigned int uint;
18 typedef long long int ll;
19 typedef unsigned long long int ull;
20 typedef double db;
21
22 using namespace std;
23
24 inline int getint()
25 {
26 int res=0;
27 char c=getchar();
28 bool mi=false;
29 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
30 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
31 return mi ? -res : res;
32 }
33 inline ll getll()
34 {
35 ll res=0;
36 char c=getchar();
37 bool mi=false;
38 while(c<'0' || c>'9') mi=(c=='-'),c=getchar();
39 while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();
40 return mi ? -res : res;
41 }
42
43 //==========================================================
44 //==========================================================
45 //==========================================================
46 //==========================================================
47
48
49 struct edge
50 {
51 int in;
52 edge*nxt;
53 }pool[40050];
54 edge*et=pool;
55 edge*eds[205];
56 void addedge(int a,int b)
57 { et->in=b; et->nxt=eds[a]; eds[a]=et++; }
58 #define FOREACH_EDGE(i,j) for(edge*i=eds[j];i;i=i->nxt)
59
60 int n,m;
61
62 char inp1[105];
63 int pos[256];
64 char inp2[1050];
65
66 int deg[105];
67
68 int cur[105];
69 bool ins[105];
70 void DFS(int dep)
71 {
72 if(dep==n)
73 {
74 for(int i=0;i<n;i++) printf("%c",inp1[cur[i]]);
75 printf("
");
76 return ;
77 }
78
79 for(int i=0;i<n;i++)
80 if(!ins[i] && deg[i]==0)
81 {
82 cur[dep]=i;
83 ins[i]=true;
84 FOREACH_EDGE(e,i) deg[e->in]--;
85 DFS(dep+1);
86 FOREACH_EDGE(e,i) deg[e->in]++;
87 ins[i]=false;
88 }
89 }
90
91 int main()
92 {
93 while(true)
94 {
95 gets(inp1);
96 if(feof(stdin)) break;
97 gets(inp2);
98
99 {
100 n=0;
101 int p=0;
102 while(inp1[p]!=0)
103 {
104 if('a'<=inp1[p] && inp1[p]<='z')
105 pos[inp1[p]]=n,inp1[n++]=inp1[p];
106 p++;
107 }
108
109 for(int i=0;i<n;i++)
110 for(int j=i+1;j<n;j++)
111 if(inp1[i]>inp1[j])
112 swap(inp1[i],inp1[j]),swap(pos[inp1[i]],pos[inp1[j]]);
113
114 }
115
116 et=pool;
117 memset(eds,0,sizeof(int)*(n+1));
118 memset(deg,0,sizeof(int)*(n+1));
119
120 {
121 int p=0;
122 int c=0;
123 char a;
124 while(inp2[p]!=0)
125 {
126 if('a'<=inp2[p] && inp2[p]<='z')
127 {
128 if(c==1)
129 {
130 addedge(pos[a],pos[inp2[p]]);
131 deg[pos[inp2[p]]]++;
132 c=0;
133 }
134 else a=inp2[p],c++;
135 }
136
137 p++;
138 }
139 }
140
141 DFS(0);
142
143 printf("
");
144 }
145
146 return 0;
147 }