HDU - 3605 : acm.hdu.edu.cn/showproblem.php?pid=3605

题目:

    有1e5的的人,小于10个的星球,每个星球都有容量,每个人也有适合的星球和不适合的星球。问所有人是否能住到星球上去。

思路:

    这道题目如果直接用常用的建图方式是不行的,因为人太多了,相应的边就很多,会TLE。那要怎么做呢,因为只有十个星球,所以有很多人的特征是相同的,所以把这1e5个点可以缩成最多1024个点。这样建图跑dinic就行了。

//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000") //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3) #include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert> using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n' #define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行
#define REP(i , j , k) for(int i = j ; i < k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = 1e9+;
const double esp = 1e-;
const double PI=acos(-1.0);
const double PHI=0.61803399; //黄金分割点
const double tPHI=0.38196601; template<typename T>
inline T read(T&x){
x=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
} /*-----------------------showtime----------------------*/ struct edge{
int u,v,cap,flag;
edge(){}
edge(int u,int v,int cap,int flag):u(u),v(v),cap(cap),flag(flag){}
}es[]; int tot,s,t;
vector<int>tab[];
int dis[],cur[];
void addedge(int u,int v,int cap){
// debug(u);
tab[u].pb(tot);
es[tot++] = edge(u,v,cap,);
tab[v].pb(tot);
es[tot++] = edge(v,u,,);
} bool bfs(){
queue<int>q; q.push(s);
memset(dis,inf,sizeof(dis));
dis[s] = ;
while(!q.empty()){
int h = q.front(); q.pop();
for(int i=; i<tab[h].size(); i++){
edge & e = es[tab[h][i]];
if(e.cap > && dis[e.v] >= inf){
dis[e.v] = dis[h] + ;
q.push(e.v);
}
}
}
return dis[t] < inf;
} int dfs(int x,int maxflow){
if(x == t || maxflow == ) return maxflow;
for(int i=cur[x] ; i<tab[x].size(); i++){
cur[x] = i;
edge & e = es[tab[x][i]];
if(dis[e.v] == dis[x] + && e.cap > ){
int flow = dfs(e.v, min(maxflow, e.cap));
if(flow){
e.cap -= flow; es[tab[x][i] ^ ].cap += flow;
return flow;
}
}
}
return ;
} int dinic(){
int ans = ;
while(bfs()){ int flow;
memset(cur,,sizeof(cur));
do{
flow = dfs(s,inf);
if(flow) ans += flow;
}while(flow); }
return ans;
}
int cnt[];
int main(){
int n,m;
while(~scanf("%d%d", &n, &m) && n+m){ tot = ;
s = , t = +m+m+;
for(int i=s; i<=t; i++)tab[i].clear();
memset(cnt,,sizeof(cnt));
for(int i=; i<=n; i++){
int tmp = ;
for(int j=; j<=m; j++){
int x; scanf("%d", &x);
tmp= tmp* + x;
}
cnt[tmp]++;
}
for(int i=; i<=; i++)
{
if(cnt[i] == )continue;
addedge(s,i,cnt[i]);
for(int j=; j<m; j++){
if((i>>j) & == ) addedge(i,+j,cnt[i]);
}
}
for(int i=; i<m; i++){
int x;scanf("%d", &x);
addedge(+i,+m+i,x);
addedge(+m+i,t,inf);
}
if(n == dinic()) puts("YES");
else puts("NO");
}
return ;
}

HDU - 3605

最新文章

  1. js ES6 多行字符串 连接字符串
  2. ThreadLocal类详解:原理、源码、用法
  3. 项目积累html标签
  4. 使用Spark分析拉勾网招聘信息(三): BMR 入门
  5. [转]LINQ之路系列博客导航
  6. ANDROID_MARS学习笔记_S02_014_GSON解析JSON串为对象
  7. POJ 1861 Network
  8. mybatis 学习笔记(4) —— 批量新增数据
  9. hibernate 简单查询
  10. 基于CAShapeLayer和贝塞尔曲线的圆形进度条动画【装载】
  11. SpringMVC+Spring+Hibernate的小样例
  12. 高效DevOps的10项实践
  13. linux logrotate配置
  14. VSCode配置eslint
  15. kettel的stream lookup报错
  16. 常见六大Web安全攻防解析
  17. Date中before和after方法的使用
  18. 浅谈C语言内存管理、内存泄露、堆栈
  19. 20165225《Java程序设计》第四周学习总结
  20. jQuery鼠标悬停

热门文章

  1. ASP.NET Core on K8S深入学习(2)部署过程解析与Dashboard
  2. Codeforces Round #192 (Div. 2) (330A) A. Cakeminator
  3. UE4中UMG与C++交互 页面文本修改
  4. S2:面向对象
  5. 利用MAVEN打包可运行jar包,包括依赖的第三方包
  6. what is the CCA?
  7. java学习笔记(中级篇)—JDK动态代理
  8. Django2.2中间件详解
  9. 8.7 day28 网络编程 socket套接字 半连接池 通信循环 粘包问题 struct模块
  10. 7.19 包 logging模块 hashlib模块 openpyxl模块 深浅拷贝