【ZOJ3933 The 16th Zhejiang University Programming ContestG】【费用流】Team Formation 最多组队条件下女

Team Formation

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

---

For an upcoming programming contest, Edward, the headmaster of Marjar University, is forming several two-man teams from students of his university.

There are n students in the university numbered from 1 to n. There are two groups of students in the university and each student belongs to exactly one of the groups. For each student in the university, he/she has a list of students whom he/she will never form a team with. For two students a and b, if b is in a's list, a is also in b's list.

Edward will only form high-performance teams. A high-performance team is a team which consists of two students from different groups who are willing to form a team. In addition, Edward wants as many as possible girls to participate in the contest. You, as Edward's secretary, need to form as many high-performance teams as possible and at the same time maximize the number of girls in the teams.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1 ≤ n ≤ 500) -- the number of students.

The second line contains a binary string of length n. If the i-th character is 0 then i-th student is in the first group, otherwise the student is in the second group.

The third line contains a binary string of length n. If the i-th character is 0 then i-th student is a girl, otherwise the student is a boy.

In the following n lines, each line starts with an integer m followed with m distinct integers, denoting the list of i-th student.

Output

For each case, output two integers a and b denoting the number of high-performance teams and the number of girls.

In the next a lines, each contains two integers x_i and y_i denoting the students in i-th team.

If there are multiple solutions, you can output any of them.

Sample Input

13101000000

Sample Output

1 21 2

---

Author: LIN, Xi
Source: The 16th Zhejiang University Programming Contest

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }const int N = 505, M = N*N*6, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;int casenum, casei;int id, ST, ED;int first[N];int w[M], c[M], cap[M], cost[M], nxt[M];int f[N];int pe[N];bool e[N];void ins(int x, int y, int cap_, int cost_){w[++id] = y;cap[id] = cap_;cost[id] = cost_;nxt[id] = first[x];first[x] = id;w[++id] = x;cap[id] = 0;cost[id] = -cost_;nxt[id] = first[y];first[y] = id;}int q[int(2e5)];int h, t;struct node{int x, dis;node(int x, int dis) :x(x), dis(dis) {}bool operator < (const node&b)const{return dis > b.dis;}};priority_queue<node>Q;void inq(int x, int cost_, int pe_){if (cost_ >= f[x])return;f[x] = cost_;pe[x] = pe_;Q.push(node(x, cost_));}bool dijkstra(){while (!Q.empty())Q.pop();MS(f, 63);MS(e, 0);cap[0] = ms63;inq(ST, 0, 0);while (!Q.empty()){int x = Q.top().x; Q.pop();if (e[x])continue; e[x] = 1;if (x == ED)return 1;for (int z = first[x]; z; z = nxt[z]){if (cap[z])inq(w[z], f[x] + cost[z], z);}}return f[ED] < ms63;}bool spfa(){MS(e, 0);MS(f, 63);cap[0] = ms63;h = t = 1e5;inq(ST, 0, 0);while (h<t){int x = q[h++]; e[x] = 0;for (int z = first[x]; z; z = nxt[z]){if (cap[z])inq(w[z], f[x] + cost[z], z);}}return f[ED]<ms63;}int edge;void MCMF(){int maxflow = 0;int mincost = 0;//while (spfa())while(dijkstra()){int flow = ms63;int x = ED;while (x != ST){gmin(flow, cap[pe[x]]);x = w[pe[x] ^ 1];}maxflow += flow;mincost += f[ED] * flow;x = ED;while (x != ST){cap[pe[x]] -= flow;cap[pe[x] ^ 1] += flow;x = w[pe[x] ^ 1];}}//跑完最小费用最大流，maxflow就是组数，mincost就是男生人数，maxflow*2-mincost就是女生人数printf("%d %d\n", maxflow, maxflow*2-mincost);for (int i = 2; i <= edge; i += 2)if (cap[i] == 0){printf("%d %d\n", w[i ^ 1], w[i]);}}char group[N];char sex[N];int n;int main(){scanf("%d", &casenum);for (casei = 1; casei <= casenum; ++casei){memset(first, 0, n + 2 << 2); id = 1;scanf("%d", &n);ST = 0;ED = n + 1;scanf("%s", group + 1);scanf("%s", sex + 1);for (int i = 1; i <= n; ++i){MS(e, 1); e[i] = 0;int g; scanf("%d", &g);while (g--){int x; scanf("%d", &x);e[x] = 0;}if (group[i] == '0')for (int j = 1; j <= n; ++j)if (e[j] && group[j] == '1'){ins(i, j, 1, 0);}}edge = id;for (int i = 1; i <= n; ++i){if (group[i] == '0')ins(ST, i, 1, sex[i]=='1');else ins(i, ED, 1, sex[i]=='1');}MCMF();}return 0;}/*【题意】有n（500）个人，这些人每个人不是属于group0就是属于group1。不是女生('0')就是男生('1')每个人有些人是不想分组的我们希望使得这n个人分尽可能多的组，在这个条件下使得女生的人数尽可能多【类型】费用流【分析】显然这题可以费用流源点向组0建边，容量为1费用为0组0的人向组1的人中不是讨厌的人建边，容量为1，费用为2-女生数组1向汇点建边，容量为1费用为0然后跑最小费用最大流即可。当然建图还可以——源点向组0建边，容量为1费用为:男生?1:0组0的人向组1的人中不是讨厌的人建边，容量为1，费用为0组1向汇点建边，容量为1费用为:男生?1:0不过没什么差别。【时间复杂度&&优化】这题很卡常数。我们可以引用spfa的SLF优化，或者引用dijkstra费用流所谓dijkstra费用流，就是用dijkstra代替spfa【数据】1004110000000000*/