HDU--1054--Strategic Game【最小点覆盖】

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链接：http://acm.hdu.edu.cn/showproblem.php?pid=1054

题意：一个熊孩子玩策略游戏，他需要用最少的士兵守卫最多的道路，如果这个顶点有士兵，则和这个点相连的所有边都会被保护，问保护所有的道路最少需要的士兵数量。

思路：这实际上就是一个最小点覆盖，二分图的最小点覆盖 == 最大匹配，这不是一个二分图，我们把n个点扩成2 * n个，把他转换为二分图，最后最大匹配再除以2就是原图的最大匹配。

Hopcroft-Karp增广模板

#include<cstring>#include<string>#include<fstream>#include<iostream>#include<iomanip>#include<cstdio>#include<cctype>#include<algorithm>#include<queue>#include<map>#include<set>#include<vector>#include<stack>#include<ctime>#include<cstdlib>#include<functional>#include<cmath>using namespace std;#define PI acos(-1.0)#define MAXN 1510#define eps 1e-7#define INF 0x3F3F3F3F      //0x7FFFFFFF#define LLINF 0x7FFFFFFFFFFFFFFF#define seed 1313131#define MOD 1000000007#define ll long long#define ull unsigned ll#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1/*二分图匹配（Hopcroft-Carp的算法）。顶点下标从0开始调用：res=MaxMatch();  Nx,Ny要初始化！！！时间复杂大为 O(V^0.5 E)适用于数据较大（点较多）的二分匹配*/struct node{    int v,next;}edge[MAXN*MAXN];int head[MAXN],cnt;int Mx[MAXN],My[MAXN],Nx,Ny;int dx[MAXN],dy[MAXN],dis;bool vis[MAXN];void add_edge(int a,int b){    edge[cnt].v = b;    edge[cnt].next = head[a];    head[a] = cnt++;}bool searchP(){    queue<int>q;    dis = INF;    memset(dx,-1,sizeof(dx));    memset(dy,-1,sizeof(dy));    for(int i = 0; i < Nx; i++)        if(Mx[i] == -1){            q.push(i);            dx[i] = 0;        }    while(!q.empty()){        int u = q.front();        q.pop();        if(dx[u] > dis)  break;        for(int i = head[u]; i != -1; i = edge[i].next){            int v = edge[i].v;            if(dy[v] == -1){                dy[v] = dx[u] + 1;                if(My[v] == -1) dis = dy[v];                else{                    dx[My[v]] = dy[v] + 1;                    q.push(My[v]);                }            }        }    }    return dis != INF;}bool DFS(int u){    for(int i = head[u]; i != -1; i = edge[i].next){        int v = edge[i].v;        if(!vis[v] && dy[v] == dx[u] + 1){            vis[v] = 1;            if(My[v] != -1 && dy[v] == dis) continue;            if(My[v] == -1 || DFS(My[v])){                My[v] = u;                Mx[u] = v;                return 1;            }        }    }    return 0;}int MaxMatch(){    int res = 0;    memset(Mx,-1,sizeof(Mx));    memset(My,-1,sizeof(My));    while(searchP()){        memset(vis,0,sizeof(vis));        for(int i = 0; i < Nx; i++)          if(Mx[i] == -1 && DFS(i))  res++;    }    return res;}int main(){    int n,i,j;    int a,m,b;    while(scanf("%d", &n)!=EOF){        cnt = 0;        memset(head,-1,sizeof(head));        for(i = 0; i < n; i++){            scanf("%d:(%d)", &a, &m);            while(m--){                scanf("%d", &b);                add_edge(a,b);                add_edge(b,a);            }        }        Nx = Ny = n;        int ans = MaxMatch();        printf("%d\n", ans / 2);    }    return 0;}

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