题目链接:点击打开链接
白书例题P369
注意的就是即使这队接下来全胜也是无法赢,要直接判掉。
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
#define ll int
#define N 700
#define M 20000
#define inf 107374182
#define inf64 1152921504606846976
struct Edge{
ll from, to, cap, nex;
}edge[M*2];//注意这个一定要够大 不然会re 还有反向弧
ll head[N], edgenum;
void add(ll u, ll v, ll cap, ll rw = 0){ //如果是有向边则:add(u,v,cap); 如果是无向边则:add(u,v,cap,cap);
Edge E = { u, v, cap, head[u]};
edge[ edgenum ] = E;
head[u] = edgenum ++;
Edge E2= { v, u, rw, head[v]};
edge[ edgenum ] = E2;
head[v] = edgenum ++;
}
ll sign[N];
bool BFS(ll from, ll to){
memset(sign, -1, sizeof(sign));
sign[from] = 0;
queue<ll>q;
q.push(from);
while( !q.empty() ){
ll u = q.front(); q.pop();
for(ll i = head[u]; i!=-1; i = edge[i].nex)
{
ll v = edge[i].to;
if(sign[v]==-1 && edge[i].cap)
{
sign[v] = sign[u] + 1, q.push(v);
if(sign[to] != -1)return true;
}
}
}
return false;
}
ll Stack[N], top, cur[N];
ll Dinic(ll from, ll to){
ll ans = 0;
while( BFS(from, to) )
{
memcpy(cur, head, sizeof(head));
ll u = from; top = 0;
while(1)
{
if(u == to)
{
ll flow = inf, loc;//loc 表示 Stack 中 cap 最小的边
for(ll i = 0; i < top; i++)
if(flow > edge[ Stack[i] ].cap)
{
flow = edge[Stack[i]].cap;
loc = i;
}
for(ll i = 0; i < top; i++)
{
edge[ Stack[i] ].cap -= flow;
edge[Stack[i]^1].cap += flow;
}
ans += flow;
top = loc;
u = edge[Stack[top]].from;
}
for(ll i = cur[u]; i!=-1; cur[u] = i = edge[i].nex)//cur[u] 表示u所在能增广的边的下标
if(edge[i].cap && (sign[u] + 1 == sign[ edge[i].to ]))break;
if(cur[u] != -1)
{
Stack[top++] = cur[u];
u = edge[ cur[u] ].to;
}
else
{
if( top == 0 )break;
sign[u] = -1;
u = edge[ Stack[--top] ].from;
}
}
}
return ans;
}
void init(){memset(head,-1,sizeof head);edgenum = 0;}
int win[30], lose[30], lef[30][30], id[30][30], tot;
int n;
vector<int>ans;
bool ok(int x){
init();
int from = 0, to = tot;
int all = win[x];
for(int i = 1; i <= n; i++)
all += lef[i][x];
int flow = 0;
for(int i = 1; i <= n; i++)
for(int j = i+1; j <= n; j++)
if(lef[i][j])
{
add(from, id[i][j], lef[i][j]);
add(id[i][j], i, inf);
add(id[i][j], j, inf);
flow += lef[i][j];
}
for(int i = 1; i <= n; i++) {
if(win[i] > all)return false;
else if(win[i] < all)
add(i, to, all - win[i]);
}
return Dinic(from, to) == flow;
}
void solve(){
for(int i = 1; i <= n; i++)
if(ok(i))
ans.push_back(i);
}
void input(){
scanf("%d", &n);
for(int i = 1; i <= n; i++)scanf("%d %d", &win[i], &lose[i]);
tot = n+1;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
scanf("%d", &lef[i][j]);
tot = n+1;
for(int i = 1; i <= n; i++)
for(int j = i+1; j <= n; j++)
id[i][j] = tot++;
}
int main(){
int T;scanf("%d",&T);
while(T--){
input();
ans.clear();
solve();
for(int i = 0; i < ans.size(); i++)
printf("%d%c", ans[i], i==(int)ans.size()-1?'\n':' ');
}
return 0;
}