传送
题面：(n)支队伍进行比赛，每支需要打的比赛数目相同。每场比赛恰好一支胜，另一支败。给出每支队伍目前胜的场数和败的场数，以及每两个队伍还剩下的比赛场数，确定所有可能得冠军的球队(获胜场数最多的得冠军，可以并列)。按照从小到大顺序给出所有可能获得冠军的队伍编号。
这题确实挺妙，关键在于把他转换成任务分配模型。
因为数据范围小，所以我们可以依次判断每个人是否能夺冠。
对于队伍(x)，先算出剩下场次全赢的总获胜场数(sum).这样就能知道其他队最多只能赢(sum-w_i)场，那么我们只要看在流量不超过(sum sum-w_i)的前提下，能否将剩余的场次分配完即可。
因此把任意两队(u,v)之间的比赛(t_{u,v})抽象成一个点，从(s)向(t_{u,v})连一条容量为剩余场次(a_{u,v})的边，再从(t_{u,v})分别向(u,v)连边，容量为无穷，因为(u,v)还要向汇点连边，有(sum-w_u)和(sum-w_v)的限制.
最后只要每次跑一遍Dinic，看最大流是否等于(sumsum a_{i,j})即可。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
#include<assert.h>
#include<ctime>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 30;
const int maxN = 1e3 + 5;
const int maxe = 1e4 + 5;
In ll read()
{
ll ans = 0;
char ch = getchar(), las = ' ';
while(!isdigit(ch)) las = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(las == '-') ans = -ans;
return ans;
}
In void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
int n, s, t, w[maxn], d[maxn], a[maxn][maxn];
struct Edge
{
int nxt, to, cap, flow;
}e[maxe];
int head[maxN], ecnt = -1;
In void addEdge(int x, int y, int w)
{
e[++ecnt] = (Edge){head[x], y, w, 0};
head[x] = ecnt;
e[++ecnt] = (Edge){head[y], x, 0, 0};
head[y] = ecnt;
}
int dis[maxN];
In bool bfs()
{
Mem(dis, 0), dis[s] = 1;
queue<int> q; q.push(s);
while(!q.empty())
{
int now = q.front(); q.pop();
for(int i = head[now], v; ~i; i = e[i].nxt)
if(e[i].cap > e[i].flow && !dis[v = e[i].to])
dis[v] = dis[now] + 1, q.push(v);
}
return dis[t];
}
int cur[maxN];
In int dfs(int now, int res)
{
if(now == t || res == 0) return res;
int flow = 0, f;
for(int& i = cur[now], v; ~i; i = e[i].nxt)
{
if(dis[v = e[i].to] == dis[now] + 1 && (f = dfs(v, min(res, e[i].cap - e[i].flow))) > 0)
{
e[i].flow += f, e[i ^ 1].flow -= f;
flow += f, res -= f;
if(res == 0) break;
}
}
return flow;
}
In int maxFlow()
{
int flow = 0;
while(bfs())
{
memcpy(cur, head, sizeof(head));
flow += dfs(s, INF);
}
return flow;
}
In bool check(int x)
{
Mem(head, -1), ecnt = -1;
int sum = w[x];
for(int i = 1; i <= n; ++i) sum += a[x][i];
for(int i = 1; i <= n; ++i) if(w[i] > sum) return 0; //及时x全赢，场数还没有i多
int flow = 0;
for(int i = 1; i <= n; ++i)
for(int j = i + 1; j <= n; ++j)
{
int id = (i - 1) * n + j + n;
addEdge(s, id, a[i][j]);
addEdge(id, i, INF), addEdge(id, j, INF);
flow += a[i][j];
}
for(int i = 1; i <= n; ++i) addEdge(i, t, sum - w[i]);
return maxFlow() == flow;
}
int main()
{
int T = read();
while(T--)
{
n = read(); s = 0, t = n * n + n + 1;
for(int i = 1; i <= n; ++i) w[i] = read(), d[i] = read();
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j) a[i][j] = read();
bool flg = 1;
for(int i = 1; i <= n; ++i) if(check(i))
{
if(flg) flg = 0;
else space;
write(i);
}
enter;
}
return 0;
}