试题库问题

链接：swust的oj竟然没有SPJ，做法不一样，a不了。

题意：中文题。

分析：网络流24题第七题。多重匹配的基础问题，模型很简单。

代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<sstream>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=1050;
const int MAX=1000000100;
const int mod=100000000;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=998244353;
const int INF=1000000010;
const double pi=acos(-1.0);
typedef double db;
typedef unsigned long long ull;
int tot,u[N],v[22*N],cap[22*N],flow[22*N],pre[22*N];
void add(int a,int b,int c,int f) {
    v[tot]=b;cap[tot]=c;flow[tot]=f;pre[tot]=u[a];u[a]=tot++;
}
bool q[2*N];
int d[2*N],dis[2*N],head[2*N];
bool bfs(int a,int b) {
    int i,l=1,r=1;
    memset(q,0,sizeof(q));
    d[1]=a;q[a]=1;dis[a]=0;
    for (;l<=r;l++)
        for (i=u[d[l]];i!=-1;i=pre[i])
        if (!q[v[i]]&&cap[i]>flow[i]) q[v[i]]=1,d[++r]=v[i],dis[v[i]]=dis[d[l]]+1;
    return q[b];
}
int dfs(int a,int b,int en) {
    if (a==en||b==0) return b;
    int ret=0,f;
    for (int& i=u[a];i!=-1;i=pre[i])
    if (dis[v[i]]==dis[a]+1&&(f=dfs(v[i],min(b,cap[i]-flow[i]),en))>0) {
        flow[i]+=f;flow[i^1]-=f;
        ret+=f;b-=f;
        if (b==0) break ;
    }
    return ret;
}
int main()
{
    int a,b,i,j,k,n,sum=0,ans=0;
    scanf("%d%d", &k, &n);
    tot=0;memset(u,-1,sizeof(u));
    for (i=1;i<=n;i++) add(0,i,1,0),add(i,0,0,0);
    for (i=1;i<=k;i++) {
        scanf("%d", &a);sum+=a;
        add(n+i,n+k+1,a,0);add(n+k+1,n+i,0,0);
    }
    for (i=1;i<=n;i++) {
        scanf("%d", &a);
        for (j=1;j<=a;j++) {
            scanf("%d", &b);
            add(i,n+b,1,0);add(n+b,i,0,0);
        }
    }
    for (i=0;i<=n+k+1;i++) head[i]=u[i];
    while (bfs(0,n+k+1)) {
        ans+=dfs(0,INF,n+k+1);
        for (i=0;i<=n+k+1;i++) u[i]=head[i];
    }
    if (sum!=ans) printf("No Solution!\n");
    else {
        for (i=1;i<=k;i++) {
            printf("%d: ", i);a=0;
            for (j=u[n+i];j!=-1;j=pre[j])
            if (flow[j]&&v[j]!=n+k+1) printf("%d ", v[j]);
            printf("\n");
        }
    }
    return 0;
}