有n個人要移居m個星球,給出每個合適的星球,每個星球最多能容納的人數,問是否所有人都可以移居。
一開始還以爲是最大流的裸題,結果TLE,翻了翻題解才知道原因:N的範圍太大,那麼多條邊連完就TLE了。之後,又WA了RE了T了幾次終於調過。
首先要明確一點,這麼多人肯定有許多人的選擇是重複的,而且總方案數不超過
我用的是二進制判重縮點。
#include<cstdio>
#include<cstring>
#include<vector>
#define MAXN 20100
using namespace std;
int d[MAXN],vd[MAXN],n,m,s,t,a,flow,cnt[MAXN];
inline int Min(int a,int b)
{return a<b?a:b;}
struct E
{
int v,w,op;
E(){}
E(int a,int b,int c)
{v = a; w = b; op = c;}
};
vector<E> g[MAXN];
int aug(int i,int augco)
{
int j,augc = augco,mind = t-1,delta,sz = g[i].size();
if(i == t) return augco;
for(j = 0; j < sz; j++)
{
int v = g[i][j].v;
if(g[i][j].w)
{
if(d[i] == d[v]+1)
{
delta = Min(g[i][j].w,augc);
delta = aug(v,delta);
g[i][j].w -= delta;
g[v][g[i][j].op].w += delta;
augc -= delta;
if(d[s] >= t) return augco - augc;
if(augc == 0) break;
}
if(d[v] < mind) mind = d[v];
}
}
if(augco == augc)
{
vd[d[i]]--;
if(vd[d[i]] == 0) d[s] = t;
d[i] = mind+1;
vd[d[i]]++;
}
return augco - augc;
}
void sap()
{
flow = 0;
memset(d,0,sizeof d);
memset(vd,0,sizeof vd);
vd[0] = t;
while(d[s] < t)
flow += aug(s,0x3f3f3f3f);
}
void init()
{
for(int i = 1; i <= t; i++)
g[i].clear();
memset(cnt,0,sizeof cnt);
}
int main()
{
while(scanf("%d%d",&n,&m) != EOF)
{
s = (1<<m); t = s+m+1;
for(int i = 1; i <= n; i++)
{
int temp = 0;
for(int j = 1; j <= m; j++)
{
scanf("%d",&a);
if(a)
temp |= 1<<(j-1);
}
cnt[temp]++;//記錄選擇爲temp的人數
}
for(int i = 1; i <= (1<<m)-1; i++)
{
if(cnt[i])
{
g[s].push_back(E(i,cnt[i],g[i].size()));
g[i].push_back(E(s,0,g[s].size()-1));
int temp = i;
for(int j = 1; (1<<(j-1)) <= temp; j++)
if(temp&1<<(j-1))//找到對應的星球
{
g[i].push_back(E(s+j,cnt[i],g[s+j].size()));
g[s+j].push_back(E(i,0,g[i].size()-1));
}
}
}
for(int i = 1; i <= m; i++)
{
scanf("%d",&a);
g[s+i].push_back(E(t,a,g[t].size()));
g[t].push_back(E(s+i,0,g[s+1].size()-1));
}
sap();
if(flow == n) printf("YES\n");
else printf("NO\n");
init();
}
}