HDU-3081-Marriage Match II

HDU-3081-Marriage Match II

http://acm.hdu.edu.cn/showproblem.php?pid=3081

女生和男生配对，有些女生相互是朋友，每个女生也可以跟她朋友所配对的男生配对

每次配对，每个女生都要跟不同的男生配对。问最多能配对几轮。

最大流，用并查集处理女生之间的朋友关系，最少配0轮，最多配n轮，二分解之，源点向女生建边，男生向汇点建边，容量均为mid，女生跟所有能配对的男生连线，容量为1，如果最大流 = mid * n，那mid就就能做到mid轮配对

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int maxn = 500;const int maxm = 50000;const int INF=0x7fffffff;int idx,n,m,p;int cur[maxn], pre[maxn];int dis[maxn], gap[maxn];int aug[maxn], head[maxn];int f[maxn];int hashh[maxn][maxn];struct Node{int u, v, w;int next;}edge[maxm];struct peo{int x;int y;}peo[maxm];void init(){int i;for(i=1;i<=n;i++)f[i]=i;}int find(int x){int r=x;while(f[r]!=r)r=f[r];f[x]=r;return f[x];}void Union(int x,int y){int xx,yy;xx=find(x);yy=find(y);if(xx!=yy)f[xx]=yy;}void addEdge(int u, int v, int w){    edge[idx].u = u;    edge[idx].v = v;    edge[idx].w = w;    edge[idx].next = head[u];    head[u] = idx++;    edge[idx].u = v;    edge[idx].v = u;    edge[idx].w = 0;    edge[idx].next = head[v];    head[v] = idx++;}void build(int mid){int i,j,st,ed,a,b;idx = 0;    memset(head, -1, sizeof(head));st=0;ed=2*n+1;for(i=1;i<=n;i++){addEdge(st,i,mid);  //源点向女孩建边addEdge(i+n,ed,mid); //男孩向汇点建边}memset(hashh,0,sizeof(hashh));for(i=0;i<m;i++){a=peo[i].x;b=peo[i].y;for(j=1;j<=n;j++){if(f[a]==f[j]&&!hashh[j][b])  //女孩的朋友及自己向男孩连边{  hashh[j][b]=1;addEdge(j,b+n,1);}}}}int SAP(int s, int e, int n)  //这个模板在网上搜的{    int max_flow = 0, v, u = s;    int id, mindis;    aug[s] = INF;    pre[s] = -1;    memset(dis, 0, sizeof(dis));    memset(gap, 0, sizeof(gap));    gap[0] = n; // 我觉得这一句要不要都行，因为dis[e]始终为0    for (int i = 0; i <= n; ++i)  // 初始化当前弧为第一条弧    cur[i] = head[i];    while (dis[s] < n)    {        bool flag = false;        if (u == e)        {            max_flow += aug[e];            for (v = pre[e]; v != -1; v = pre[v]) // 路径回溯更新残留网络            {                id = cur[v];                edge[id].w -= aug[e];                edge[id^1].w += aug[e];                aug[v] -= aug[e]; // 修改可增广量，以后会用到                if (edge[id].w == 0) u = v; // 不回退到源点，仅回退到容量为0的弧的弧尾            }        }        for (id = cur[u]; id != -1; id = edge[id].next)        {   // 从当前弧开始查找允许弧            v = edge[id].v;            if (edge[id].w > 0 && dis[u] == dis[v] + 1) // 找到允许弧            {                flag = true;                pre[v] = u;                cur[u] = id;                aug[v] = min(aug[u], edge[id].w);                u = v;                break;            }        }        if (flag == false)        {            if (--gap[dis[u]] == 0) break; /* gap优化，层次树出现断层则结束算法 */            mindis = n;            cur[u] = head[u];            for (id = head[u]; id != -1; id = edge[id].next)            {                v = edge[id].v;                if (edge[id].w > 0 && dis[v] < mindis)                {                    mindis = dis[v];                    cur[u] = id; // 修改标号的同时修改当前弧                }            }            dis[u] = mindis + 1;            gap[dis[u]]++;            if (u != s) u = pre[u]; // 回溯继续寻找允许弧        }    }    return max_flow;}int main(){    int i,t,ans;int a,b;int low,high,mid;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d",&n,&m,&p);for(i=0;i<m;i++)scanf("%d%d",&peo[i].x,&peo[i].y);init();while(p--){scanf("%d%d",&a,&b);Union(a,b);}for(i=1;i<=n;i++)        f[i]=find(i);        low=0;high=n;  //最多n轮ans=0;while(low<=high)   //二分求解{mid=(low+high)/2;build(mid);if(SAP(0,2*n+1,2*n+2)==n*mid)  //可以进行mid轮{low=mid+1;ans=mid;}elsehigh=mid-1;}printf("%d\n",ans);}system("pause");return 0;}