标签：hdu mat http str 超级 line tar cond 分享图片
目录
（有任何问题欢迎留言或私聊 && 欢迎交流讨论哦
?原题目描述在最下面。
?\(n(n\leq 100000)\)个人\(m(m\leq 10)\)个星球，每个星球容纳的人数有上限，每个人都有自己想去的星球意愿。问是否所有的都能去到外星球。
?肯定不能暴力建图，因为人太多了。注意到m的大小只有10。刷过状压dp的人应该都能想到要二进制优化。
?这是每个人其实都是没有区别的，有区别的只是他们的意愿。然后最多也只有\(2^{10} =1024\)种意愿。那就把这些当作\(1000\)多个点嘛。记录每种状态的人数。
细节见代码。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#define mme(a,b) memset((a),(b),sizeof((a)))
#define fuck(x) cout<<"* "<<x<<"\n"
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int MXN = 1e4+7;
const int MXE = 1e7+7;
struct DINIC{
int tot,vt,vs;
int d[MXN],head[MXN];
struct lp{
int to,w,nex;
}cw[MXE];
void add_edge(int a,int b,int c){
cw[++tot].to=b;cw[tot].nex=head[a],cw[tot].w=c;
head[a]=tot;
cw[++tot].to=a;cw[tot].nex=head[b],cw[tot].w=0;
head[b]=tot;
}
bool bfs(){
memset(d,-1,sizeof(d));
queue<int>Q;
Q.push(vt);d[vt]=0;
while(!Q.empty()){
int u=Q.front();
Q.pop();
for(int i=head[u];i!=-1;i=cw[i].nex){
int v=cw[i].to;
if(cw[i^1].w&&d[v]==-1){
d[v]=d[u]+1;
Q.push(v);
}
}
}
return d[vs]!=-1;
}
int dfs(int x,int f){
if(x==vt||f==0) return f;
int use=0,w;
for(int i=head[x];i!=-1;i=cw[i].nex){
int to=cw[i].to;
if(d[to]==d[x]-1 && cw[i].w){
w=dfs(to,min(cw[i].w,f-use));
cw[i].w-=w,cw[i^1].w+=w;
use+=w;
if(use==f) return f;
}
}
return use;
}
void init(int st,int ed){
tot=-1;
memset(head,-1,sizeof(head));
vs=st;vt=ed;
}
int max_flow(){
int ans=0;
while(bfs())ans+=dfs(vs,INF);
return ans;
}
}dinic;
int n, m;
int vs, vt;
int num[1<<11];
int main(){
while(~scanf("%d%d",&n,&m)){
int state = 1<<m;
mme(num,0);
vs=state+m+1,vt=state+m+2;
dinic.init(vs,vt);
for(int i=1;i<=n;++i){
int tmp=0;
for(int j=1;j<=m;++j){
int x;scanf("%d",&x);
if(x)tmp|=(1<<(j-1));
}
num[tmp]++;
}
for(int i=1;i<=m;++i){
int x;scanf("%d",&x);
dinic.add_edge(vs,i+state,x);
}
for(int i=1;i<state;++i){
if(num[i]==0)continue;
dinic.add_edge(i,vt,num[i]);
for(int j=0;j<m;++j){
if(i&(1<<j)){
dinic.add_edge(state+j+1,i,num[i]);
}
}
}
LL ans = dinic.max_flow();
if(ans>=n)printf("YES\n");
else printf("NO\n");
}
return 0;
}
Problem Description
2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.
Input
More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000
Output
Determine whether all people can live up to these stars
If you can output YES, otherwise output NO.
Sample Input
1 1
1
1
2 2
1 0
1 0
1 1
Sample Output
YES
NO
Source
2010 ACM-ICPC Multi-University Training Contest（17）——Host by ZSTU
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标签：hdu mat http str 超级 line tar cond 分享图片
原文地址：https://www.cnblogs.com/Cwolf9/p/9418994.html
