圆桌问题 【网络流24题】

 

 

 

 

 

 

思路

　　题目已经给出了暗示

　　即题目中给出了一个二分图

　　显然，可以把人和桌子作为二分图的两个阵营

　　由于题目中给出的限制：每个单位的人都必须坐在不同的桌子上

　　所以把单位向每个桌子连一条流量为1的边

　　由S向单位连单位人数的边　

　　由桌子向T连桌子容量的边　　

　　然后跑一次最大流，最后枚举每个单位的邻接边，通过判断残量网络上的值可以求出这个人去了哪张桌子

　　朴素的Dinic会T最后一个点

　　考虑当前弧优化来优化朴素Dinic算法，只需要70ms即可跑完

 

CODE

 

#include <bits/stdc++.h>
#define dbg(x) cout << #x << "=" << x << endl
#define eps 1e-8
#define pi acos(-1.0)

using namespace std;
typedef long long LL;

const int inf = 0x3f3f3f3f;

template<class T>inline void read(T &res)
{
    char c;T flag=1;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
    while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
}

namespace _buff {
    const size_t BUFF = 1 << 19;
    char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
    char getc() {
        if (ib == ie) {
            ib = ibuf;
            ie = ibuf + fread(ibuf, 1, BUFF, stdin);
        }
        return ib == ie ? -1 : *ib++;
    }
}

int qread() {
    using namespace _buff;
    int ret = 0;
    bool pos = true;
    char c = getc();
    for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
        assert(~c);
    }
    if (c == '-') {
        pos = false;
        c = getc();
    }
    for (; c >= '0' && c <= '9'; c = getc()) {
        ret = (ret << 3) + (ret << 1) + (c ^ 48);
    }
    return pos ? ret : -ret;
}

const int maxn = 5e4 + 7;

int s, t, cnt = 1;
int head[maxn << 1], edge[maxn << 1], nxt[maxn << 1], vis[maxn << 1];
int w[maxn << 1];//残量网络
int rev[maxn << 1];
int depth[maxn << 1];//图层
int n, m;
int cur[maxn << 1];//当前弧优化

void BuildGraph(int u, int v, int cap) {
    ++cnt;
    edge[cnt] = v;
    nxt[cnt] = head[u];
    rev[cnt] = cnt + 1;
    w[cnt] = cap;
    head[u] = cnt;

    ++cnt;
    edge[cnt] = u;
    nxt[cnt] = head[v];
    w[cnt] = 0;
    rev[cnt] = cnt - 1;
    head[v] = cnt;
}

bool bfs(int x) {
    queue<int> q;
    for ( int i = 0; i <= t; ++i ) {
        cur[i] = head[i];
    }
    while(!q.empty()) {
        q.pop();
    }
    memset(depth, 63, sizeof(depth));
    depth[s] = 0;
    q.push(s);
    do {
        int u = q.front();
        q.pop();
        for ( int i = head[u]; i; i = nxt[i] ) {
            int v = edge[i];
            if(w[i] > 0 && depth[u] + 1 < depth[v]) {
                depth[v] = depth[u] + 1;
                q.push(v);
                if(v == t) {
                    return true;
                }
            }
        }
    } while (!q.empty());
    if(depth[t] < inf)
        return true;
    return false;
}

int dfs(int u, int dist) {
    if(!dist || u == t) {
        return dist;
    }
    int flow = 0, f;
    for ( int i = cur[u]; i; i = nxt[i] ) {
        cur[u] = 1;
        int v = edge[i];
        if(depth[v] == depth[u] + 1 && (f = dfs(v, min(dist, w[i])))) {
            flow += f;
            dist -= f;
            w[i] -= f;
            w[i ^ 1] += f;
            if(!dist)
                break;
        }
    }
    return flow;
}

int pp[maxn];
int table[maxn];
int sum = 0;

int main()
{
    //freopen("data.txt", "r", stdin);
    read(m); read(n);
    for ( int i = 1; i <= m; ++i ) {
        read(pp[i]);
        sum += pp[i];
    }
    for ( int i = 1; i <= n; ++i ) {
        read(table[i]);
    }
    s = 0, t = n + m + 1;
    for ( int i = 1; i <= m; ++i ) {
        BuildGraph(s, i, pp[i]);
        for ( int j = 1; j <= n; ++j ) {
            BuildGraph(i, j + m, 1);
        }
    }
    for ( int i = 1; i <= n; ++i ) {
        //printf("table[%d]:%d\n",i, table[i]);
        BuildGraph(i + m, t, table[i]);
    }   
    //dbg(sum);
    int ans = 0;
    while(bfs(s)) {
        ans += dfs(0, 0x7f7f7f7f);
    }
    //dbg(ans);
    if(sum != ans) {
        puts("0");
    }
    else {
        cout << 1 << endl;
        for ( int i = 1; i <= m; ++i ) {
            for ( int j = head[i]; j; j = nxt[j] ) {
                int to = edge[j];
                if(to != t && !w[j]) {
                    printf("%d ",to - m);
                    //printf("cap[%d]:%d\n",j, w[j]);
                }
            }
            puts("");
        }
    }
    return 0;
}