UVA 1306 LA 2531 The K-League

题意：有n只队伍打比赛 给出每只队目前获胜和失败的场数 再给出两两队伍接下来的比赛场次 问你哪些队伍可能的冠军

思路：主要考察建图方法 我们要依次判断每只队伍是否可能获胜 对于队伍i 首先预处理出i获胜的总场数sum 对每两只队伍进行编号（比如(u, v)编号为x） 从源点连一条到该编号的边 容量为这两只队伍接下来要比的场次 然后从该编号向这两只队伍连边 容量INF 最后 对于每只队伍 分别向汇点连边 容量为为sum-win[u] 其中win[u]代表u这只队伍已经赢的场次 然后判断从源点连出的边是否都满流即可

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define REP( i, a, b ) for( int i = a; i < b; i++ )
#define FOR( i, a, b ) for( int i = a; i <= b; i++ )
#define CLR( a, x ) memset( a, x, sizeof a )
#define CPY( a, x ) memcpy( a, x, sizeof a )

const int maxn = 500 + 10;
const int maxe = 250000 + 10;
const int INF = 1e9;

struct Edge{
int v, c, f;
int next;
Edge() {}
Edge(int v, int c, int f, int next) : v(v), c(c), f(f), next(next) {}
};
struct ISAP{
int n, s, t;
int num[maxn], cur[maxn], d[maxn], p[maxn];
int Head[maxn], cntE;
int Q[maxn], head, tail;
Edge edge[maxe];
void Init(int n){
this -> n = n;
cntE = 0;
CLR(Head, -1);
}
void Add(int u, int v, int c){
edge[cntE] = Edge(v, c, 0, Head[u]);
Head[u] = cntE++;
edge[cntE] = Edge(u, 0, 0, Head[v]);
Head[v] = cntE++;
}
void Bfs(){
CLR(d, -1);
CLR(num, 0);
d[t] = 0;
head = tail = 0;
Q[tail++] = t;
num[0] = 1;
while(head != tail){
int u = Q[head++];
for(int i = Head[u]; ~i; i = edge[i].next){
Edge &e = edge[i];
if(~d[e.v]) continue;
d[e.v] = d[u] + 1;
Q[tail++] = e.v;
num[d[e.v]] ++;
}
}
}
int Maxflow(int s, int t){
this -> s = s;
this -> t = t;
CPY(cur, Head);
Bfs();
int u = p[s] = s, flow = 0;
while(d[s] < n){
if(u == t){
int f = INF, neck;
for(int i = s; i != t; i = edge[cur[i]].v){
if(f > edge[cur[i]].c - edge[cur[i]].f){
f = edge[cur[i]].c - edge[cur[i]].f;
neck = i;
}
}
for(int i = s; i != t; i = edge[cur[i]].v){
edge[cur[i]].f += f;
edge[cur[i]^1].f -= f;
}
flow += f;
u = neck;
}
int ok = 0;
for(int i = cur[u]; ~i; i = edge[i].next){
Edge &e = edge[i];
if(e.c > e.f && d[e.v] + 1 == d[u]){
ok = 1;
cur[u] = i;
p[e.v] = u;
u = e.v;
break;
}
}
if(!ok){
int m = n - 1;
if(--num[d[u]] == 0) break;
for(int i = Head[u]; ~i; i = edge[i].next){
Edge &e = edge[i];
if(e.c - e.f > 0 && m > d[e.v]){
cur[u] = i;
m = d[e.v];
}
}
++num[d[u] = m + 1];
u = p[u];
}
}
return flow;
}

}solver;

int n, win[maxn], match[maxn][maxn], S, T;

void input(){
int x;
scanf("%d", &n);
FOR(i, 1, n) scanf("%d%d", &win[i], &x);
FOR(i, 1, n) FOR(j, 1, n) scanf("%d", &match[i][j]);
}

bool build(int cur){
int sum = win[cur], flow = 0, cnt_match = n + 1, S = 0, T = n + n * (n - 1) / 2 + 1;
FOR(i, 1, n) sum += match[cur][i];
solver.Init(T + 1);
FOR(i, 1, n) FOR(j, i + 1, n){
if(sum < win[i] || sum < win[j]) return false;
//这里要注意 如果某支队伍已经赢的场次大于cur这只队伍得冠军要赢的场次 那么直接返回false
flow += match[i][j];
solver.Add(S, cnt_match, match[i][j]);
solver.Add(cnt_match, i, INF);
solver.Add(cnt_match, j, INF);
cnt_match++;
}
FOR(i, 1, n) solver.Add(i, T, sum - win[i]);
return flow == solver.Maxflow(S, T);
}

void solve(){
bool flag = false;
FOR(i, 1, n) if(build(i)){
if(flag) printf(" ");
printf("%d", i);
flag = true;
}
printf("\n");
}

int main()
{
//freopen("in.txt", "r", stdin);
int T;
scanf("%d", &T);
while(T--){
input();
solve();
}
return 0;
}