LOJ 6004 【网络流24题】圆桌聚餐

题面

题解

1. 源点向每个单位连一条边，边权是该单位人数；
2. 每个单位向各个餐桌连一条边，边权是1；
3. 每个餐桌向汇点连一条边，边权是餐桌容纳人数。

如果最大流 < 总人数，则无解。若有解，根据单位和餐桌之间哪些边跑满了来输出具体方案即可。

我犯过的错误：源点汇点的编号标错了……

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
#define space putchar(' ')
#define enter putchar('\n')
template <class T>
bool read(T &x){
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-') op = 1;
else if(c == EOF) return 0;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
return 1;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}

const int N = 4005, M = 2000005, INF = 0x3f3f3f3f;
int ncnt, n, m, s, t, ans, sum;
int ecnt = 1, adj
, nxt[M], go[M], cap[M], cur
;
int que
, qr, lev
, stk
, top;

void ADD(int u, int v, int w){
go[++ecnt] = v;
nxt[ecnt] = adj[u];
adj[u] = ecnt;
cap[ecnt] = w;
}
void add(int u, int v, int w){
ADD(u, v, w), ADD(v, u, 0);
}
bool bfs(){
for(int i = 1; i <= ncnt; i++)
lev[i] = -1, cur[i] = adj[i];
lev[s] = 0, que[qr = 1] = s;
for(int ql = 1; ql <= qr; ql++){
int u = que[ql];
for(int e = adj[u], v; e; e = nxt[e])
if(cap[e] && lev[v = go[e]] == -1){
lev[v] = lev[u] + 1, que[++qr] = v;
if(v == t) return 1;
}
}
return 0;
}
int dinic(int u, int flow){
if(u == t) return flow;
int delta, ret = 0;
for(int &e = cur[u], v; e; e = nxt[e])
if(cap[e] && lev[v = go[e]] > lev[u]){
delta = dinic(v, min(cap[e], flow - ret));
if(delta){
cap[e] -= delta;
cap[e ^ 1] += delta;
ret += delta;
if(ret == flow) return flow;
}
}
lev[u] = -1;
return ret;
}

int main(){
read(m), read(n);
ncnt = n + m + 2, s = ncnt - 1, t = ncnt;
for(int i = 1, val; i <= m; i++)
read(val), add(s, i, val), sum += val;
for(int i = 1, val; i <= n; i++)
read(val), add(i + m, t, val);
for(int i = 1; i <= m; i++)
for(int j = 1; j <= n; j++)
add(i, j + m, 1);
while(bfs()) ans += dinic(s, INF);
if(ans < sum){
puts("0");
return 0;
}
puts("1");
for(int u = 1; u <= m; u++){
for(int e = adj[u]; e; e = nxt[e])
if(!(e & 1) && !cap[e])
write(go[e] - m), space;
enter;
}
return 0;
}