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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#567919 | #9315. Rainbow Bracket Sequence | Lavender_Field | TL | 0ms | 3780kb | C++20 | 2.9kb | 2024-09-16 14:39:00 | 2024-09-16 14:39:00 |
Judging History
answer
#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(Int i=a;i>=b;i--)
#define pb push_back
typedef long long ll;
using namespace std;
inline int rd() {
int r = 0; bool w = false; char ch = getchar();
while( ch < '0' || ch > '9' ) w = !(ch^45), ch = getchar();
while( ch >= '0' && ch <= '9' ) r = (r<<1) + (r<<3) + (ch^48), ch = getchar();
return w ? -r : r;
}
#define MAXN 100
#define MAXM 100
#define MAXE 20000
#define MAXD 500
const int INF = 1e9;
int n, m, s, t, tr, ans, ans_edg;
ll cost, total;
int lim[MAXM+5], v[MAXN*2+5], c[MAXN*2+5];
struct Edge
{
int fwd,nxt,wht;
ll dis;
}edg[MAXE+5];
int head[MAXD+5],edg_num=1;
inline void add( int u , int v , int w , int d )
{
++edg_num;
edg[edg_num].fwd=v;
edg[edg_num].wht=w;
edg[edg_num].dis=d;
edg[edg_num].nxt=head[u];
head[u]=edg_num;
}
inline void adde( int u , int v , int w , int d ) {
add(u,v,w,d);
add(v,u,0,-d);
}
vector<int> chk_max;
ll dist[MAXD+5];
int flow[MAXD+5],pre[MAXD+5];
bool vis[MAXD+5];
queue<int> q;
inline bool SPFA()
{
memset(dist,0x3c,sizeof(dist));
memset(vis,false,sizeof(vis));
while( !q.empty() ) q.pop();
q.push(s),dist[s]=0,flow[s]=INF;
while( !q.empty() )
{
int u=q.front();q.pop();
vis[u]=false;
for(int i=head[u];i;i=edg[i].nxt) if( edg[i].wht )
{
int v=edg[i].fwd;
if( dist[v]>dist[u]+edg[i].dis )
{
dist[v]=dist[u]+edg[i].dis;
pre[v]=i;
flow[v]=min(flow[u],edg[i].wht);
if( !vis[v] )
q.push(v),vis[v]=true;
}
}
}
return dist[t]!=dist[0];
}
inline void MCMF()
{
while( SPFA() )
{
ans+=flow[t],cost+=dist[t]*flow[t];
int p=pre[t];
while( p )
{
edg[p].wht-=flow[t],edg[p^1].wht+=flow[t];
p=pre[edg[p^1].fwd];
}
}
}
void solve() {
edg_num = 1;
memset(head, 0, sizeof(head));
chk_max.clear();
n = rd(), m = rd();
s = 3 * n + m + 1; tr = s + 1; t = tr + 1;
cost = ans = 0;
total = (ll)(1e9+1) * n;
FOR(i,1,m) lim[i] = rd();
FOR(i,1,2*n) c[i] = rd();
FOR(i,1,2*n) v[i] = rd();
FOR(i,1,n) adde(s,i,1,0);
FOR(i,1,n) {
FOR(j,i,2*i-1)
adde(i,n+j,1,0);
}
FOR(i,1,2*n) {
adde(n+i,3*n+c[i],1,1e9+1-v[i]);
}
FOR(i,1,m) {
adde(s, tr, lim[i], 0);
chk_max.pb(edg_num-1);
adde(3*n+i, t, lim[i], 0);
chk_max.pb(edg_num-1);
adde(3*n+i, tr, n, 0);
}
adde(tr, t, n, 0);
ans_edg = edg_num-1;
MCMF();
if( edg[ans_edg].wht != 0 ) {
// printf("%d\n", ans);
puts("-1");
return;
}
for(auto e: chk_max) if( edg[e].wht != 0 ) {
// puts("edg-1");
puts("-1");
return;
}
printf("%lld\n", total - cost);
}
int main() {
int T = rd(); while( T-- ) solve();
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 0ms
memory: 3780kb
input:
2 3 2 1 2 1 2 2 2 1 2 3 1 4 2 2 1 3 2 2 2 1 2 2 2 1 2 3 1 4 2 2 1
output:
9 -1
result:
ok 2 number(s): "9 -1"
Test #2:
score: -100
Time Limit Exceeded
input:
50 8 6 0 2 2 0 3 1 3 5 1 1 5 3 5 2 3 2 2 1 2 2 4 6 998133227 879226371 59632864 356493387 62611196 827258251 296576565 204244054 812713672 780267148 614679390 447700005 102067050 544546349 116002772 761999375 1 1 1 1 1 343766215 374461155 3 1 2 1 1 1 1 1 1 796323508 303640854 701432076 853325162 610...