1191: [HNOI2006]超级英雄Hero

1191: [HNOI2006]超级英雄Hero

链接

分析：

　　二分+网络流，二分答案，网络流判断。

　　或者匈牙利，从1到n挨个匹配。

匈牙利：0ms

#include<cstdio>
#include<cstring>
#include<cctype>

const int N = 1010;

struct Edge{
    int to,nxt;
    Edge() {}
    Edge(int a,int b) {to = a,nxt = b;}
}e[1000100];
int head[N],match[N],tot;
bool vis[N];
int n,m;

inline int read() {
    int x = 0,f = 1;char ch = getchar();
    for (; !isdigit(ch); ch=getchar()) if(ch=='-') f=-1;
    for (; isdigit(ch); ch=getchar()) x=x*10+ch-'0';
    return x * f;
}
bool dfs(int u) {
    for (int i=head[u]; i; i=e[i].nxt) {
        int v = e[i].to - m;
        if (!vis[v]) {
            vis[v] = true;
            if (!match[v] || dfs(match[v])) {
                match[v] = u;
                return true;
            }
        }
    }
    return false;
}
int main () {
    n = read(),m = read();
    for (int i=1; i<=m; ++i) {
        int x = read() + 1,y = read() + 1;
        e[++tot] = Edge(x+m,head[i]);head[i] = tot;
        e[++tot] = Edge(y+m,head[i]);head[i] = tot;
    }
    int ans = 0;
    for (int i=1; i<=m; ++i) {
        memset(vis,false,sizeof(vis));
        if (dfs(i)) ans++;
        else break;
    }
    printf("%d",ans);
    return 0;
}

网络流：72ms

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cctype>
 
using namespace std;
 
const int N = 5010;
 
inline int read() {
    int x = 0,f = 1;char ch=getchar();
    for (; !isdigit(ch); ch=getchar()) if(ch=='-')f=-1;
    for (; isdigit(ch); ch=getchar()) x=x*10+ch-'0';
    return x*f;
}
struct Edge{
    int to,nxt,c;
    Edge() {}
    Edge(int x,int y,int z) {to = x,c = y,nxt = z;}
}e[100010];
int head[N],dis[N],cur[N],q[N],a[N],b[N];
int L,R,tot,S,T,n,m;
 
void add_edge(int u,int v,int w) {
    e[++tot] = Edge(v,w,head[u]);head[u] = tot;
    e[++tot] = Edge(u,0,head[v]);head[v] = tot;
}
bool bfs() {
    for (int i=1; i<=T; ++i) cur[i] = head[i],dis[i] = -1;
    L = 1;R = 0;
    q[++R] = S;
    dis[S] = 0;
    while (L <= R) {
        int u = q[L++];
        for (int i=head[u]; i; i=e[i].nxt) {
            int v = e[i].to;
            if (dis[v] == -1 && e[i].c > 0) {
                dis[v] = dis[u] + 1;
                q[++R] = v;
                if (v == T) return true;
            }
        }
    }
    return false;
}
int dfs(int u,int flow) {
    if (u == T) return flow;
    int used = 0,tmp;
    for (int i=cur[u]; i; i=e[i].nxt) {
        int v = e[i].to;
        if (dis[v] == dis[u] + 1 && e[i].c > 0) {
            tmp = dfs(v,min(flow-used,e[i].c));
            if (tmp) {
                e[i].c -= tmp;e[i^1].c += tmp;
                used += tmp;
                if (used == flow) break;
            }
        }
    }
    if (used != flow) dis[u] = -1;
    return used;
}
bool check(int x) {
    memset(head,0,sizeof(head));
    tot = 1;
    S = x + n + 1,T = x + n + 2;
    for (int i=1; i<=x; ++i) {   
        add_edge(i+n,a[i],1);
        add_edge(i+n,b[i],1);   
        add_edge(S,i+n,1);      
    }
    for (int i=1; i<=n; ++i) add_edge(i,T,1);
    int ans = 0;
    while (bfs()) 
    ans += dfs(S,1e9);
    return ans==x;
}
int main() {
    n = read(),m = read();
    for (int i=1; i<=m; ++i) {
        a[i] = read()+1,b[i] = read()+1;
    }
    int L = 1,R = m,ans = 1;
    while (L <= R) {
        int mid = (L + R) / 2;
        if (check(mid)) ans = mid,L = mid + 1;
        else R = mid - 1;
    }
    cout << ans;
    return 0;
}