POJ 2771 Guardian of Decency (二分圖最大點獨立集)
阿新 • • 發佈:2018-03-13
str number iostream hit school ear one ability clas
So, for any two persons that he brings on the excursion, they must satisfy at least one of the requirements above. Help him find the maximum number of persons he can take, given their vital information.
The first line of the input consists of an integer T ≤ 100 giving the number of test cases. The first line of each test case consists of an integer N ≤ 500 giving the number of pupils. Next there will be one line for each pupil consisting of four space-separated data items:
No string in the input will contain more than 100 characters, nor will any string contain any whitespace.
Guardian of Decency
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 6133 | Accepted: 2555 |
Description
Frank N. Stein is a very conservative high-school teacher. He wants to take some of his students on an excursion, but he is afraid that some of them might become couples. While you can never exclude this possibility, he has made some rules that he thinks indicates a low probability two persons will become a couple:- Their height differs by more than 40 cm.
- They are of the same sex.
- Their preferred music style is different.
- Their favourite sport is the same (they are likely to be fans of different teams and that would result in fighting).
So, for any two persons that he brings on the excursion, they must satisfy at least one of the requirements above. Help him find the maximum number of persons he can take, given their vital information.
Input
- an integer h giving the height in cm;
- a character ‘F‘ for female or ‘M‘ for male;
- a string describing the preferred music style;
- a string with the name of the favourite sport.
No string in the input will contain more than 100 characters, nor will any string contain any whitespace.
Output
For each test case in the input there should be one line with an integer giving the maximum number of eligible pupils.Sample Input
2 4 35 M classicism programming 0 M baroque skiing 43 M baroque chess 30 F baroque soccer 8 27 M romance programming 194 F baroque programming 67 M baroque ping-pong 51 M classicism programming 80 M classicism Paintball 35 M baroque ping-pong 39 F romance ping-pong 110 M romance Paintball
Sample Output
3 7
題意
一共n個人,從中選出最多的人去旅遊。當兩個人可能成為couples時兩個人不能同時去,可能成為couples的條件是:1、身高差小於等於40;2、不同性別;3、喜歡相同的音樂;4、喜歡不同的運動。
分析
如果兩個人可能成為couples,連邊,求最大點獨立集。
二分圖的最大點獨立集 = n-最小點覆蓋集=n-最大匹配數。
code
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<string> 5 #include<iostream> 6 7 using namespace std; 8 const int N = 1010; 9 const int INF = 1e9; 10 11 struct Student{ 12 int h; 13 string x,y,s; 14 }data[510]; 15 struct Edge{ 16 int to,nxt,c; 17 Edge() {} 18 Edge(int x,int y,int z) {to = x,c = y,nxt = z;} 19 }e[1000100]; 20 int q[1000100],L,R,S,T,tot = 1; 21 int dis[N],cur[N],head[N]; 22 23 void add_edge(int u,int v,int c) { 24 e[++tot] = Edge(v,c,head[u]);head[u] = tot; 25 e[++tot] = Edge(u,0,head[v]);head[v] = tot; 26 } 27 bool bfs() { 28 for (int i=1; i<=T; ++i) cur[i] = head[i],dis[i] = -1; 29 L = 1,R = 0; 30 q[++R] = S;dis[S] = 1; 31 while (L <= R) { 32 int u = q[L++]; 33 for (int i=head[u]; i; i=e[i].nxt) { 34 int v = e[i].to; 35 if (dis[v] == -1 && e[i].c > 0) { 36 dis[v] = dis[u]+1;q[++R] = v; 37 if (v==T) return true; 38 } 39 } 40 } 41 return false; 42 } 43 int dfs(int u,int flow) { 44 if (u==T) return flow; 45 int used = 0; 46 for (int &i=cur[u]; i; i=e[i].nxt) { 47 int v = e[i].to; 48 if (dis[v] == dis[u] + 1 && e[i].c > 0) { 49 int tmp = dfs(v,min(flow-used,e[i].c)); 50 if (tmp > 0) { 51 e[i].c -= tmp;e[i^1].c += tmp; 52 used += tmp; 53 if (used == flow) break; 54 } 55 } 56 } 57 if (used != flow) dis[u] = -1; 58 return used; 59 } 60 int dinic() { 61 int ret = 0; 62 while (bfs()) ret += dfs(S,INF); 63 return ret; 64 } 65 void Clear() { 66 tot = 1; 67 memset(head,0,sizeof(head)); 68 } 69 int main() { 70 int Case,n; 71 scanf("%d",&Case); 72 while (Case--) { 73 Clear(); 74 scanf("%d",&n); 75 S = n+n+1;T = n+n+2; //- 76 for (int i=1; i<=n; ++i) { 77 scanf("%d",&data[i].h); 78 cin >> data[i].x >> data[i].y >> data[i].s; 79 } 80 for (int i=1; i<=n; ++i) 81 for (int j=1; j<=n; ++j) { // 可能成為couples的連邊 82 if (abs(data[i].h-data[j].h) > 40) continue; 83 if (data[i].x == data[j].x) continue; 84 if (data[i].y != data[j].y) continue; 85 if (data[i].s == data[j].s) continue; 86 add_edge(i,j+n,1); 87 } 88 for (int i=1; i<=n; ++i) add_edge(S,i,1),add_edge(i+n,T,1); 89 int ans = dinic(); 90 printf("%d\n",n-ans/2); 91 } 92 return 0; 93 }
POJ 2771 Guardian of Decency (二分圖最大點獨立集)