贴个板子

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N=205, M=5e5+5, INF=1e9;
inline ll read(){
char c=getchar();ll x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
} int n, m, u, v, c, b, s, t;
int extra[N];
struct edge{int v, c, f, ne, lower;}e[M];
int cnt=1, h[N];
inline void ins(int u, int v, int c, int b=0) { //printf("ins %d %d %d\n",u,v,c);
e[++cnt]=(edge){v, c, 0, h[u], b}; h[u]=cnt;
e[++cnt]=(edge){u, 0, 0, h[v], b}; h[v]=cnt;
} int q[N], head, tail, d[N], vis[N], cur[N];
bool bfs() {
memset(vis, 0, sizeof(vis));
head=tail=1;
q[tail++]=s; d[s]=0; vis[s]=1;
while(head!=tail) {
int u=q[head++];
for(int i=h[u];i;i=e[i].ne)
if(!vis[e[i].v] && e[i].c > e[i].f) {
vis[e[i].v]=1; d[e[i].v]=d[u]+1;
q[tail++]=e[i].v;
if(e[i].v == t) return true;
}
}
return false;
}
int dfs(int u, int a) {
if(u==t || a==0) return a;
int flow=0, f;
for(int &i=cur[u];i;i=e[i].ne)
if(d[e[i].v] == d[u]+1 && (f=dfs(e[i].v, min(e[i].c-e[i].f, a))) >0) {
flow += f;
e[i].f += f;
e[i^1].f -= f;
a -= f;
if(a==0) break;
}
if(a) d[u]=-1;
return flow;
}
int dinic() {
int flow=0;
while(bfs()) {
for(int i=s; i<=t; i++) cur[i]=h[i];
flow+=dfs(s, INF);
}
return flow;
}
int main() {
freopen("in","r",stdin);
int T=read();
while(T--) {
n=read(); m=read(); s=0; t=n+1;
cnt=1; memset(h,0,sizeof(h)); memset(extra, 0, sizeof(extra));
for(int i=1; i<=m; i++)
u=read(), v=read(), b=read(), c=read(), ins(u, v, c-b, b), extra[u]-=b, extra[v]+=b;
int sum=0;
for(int i=1; i<=n; i++) {
if(extra[i]>0) ins(s, i, extra[i]), sum+=extra[i];
else ins(i, t, -extra[i]);
}
int flow=dinic(); //printf("flow %d %d\n",flow,sum);
if(flow!=sum) puts("NO");
else {
puts("YES");
for(int i=1; i<=m; i++) printf("%d\n",e[i<<1].f + e[i<<1].lower);
}
}
}

ZOJ 2314 Reactor Cooling [无源汇上下界网络流]的更多相关文章

  1. ZOJ 2314 - Reactor Cooling - [无源汇上下界可行流]

    题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2314 The terrorist group leaded by ...

  2. ZOJ2314 Reactor Cooling(无源汇上下界可行流)

    The terrorist group leaded by a well known international terrorist Ben Bladen is buliding a nuclear ...

  3. ZOJ 2314 Reactor Cooling(无源汇有上下界可行流)

    题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2314 题目大意: 给n个点,及m根pipe,每根pipe用来流躺 ...

  4. ZOJ 2314 Reactor Cooling | 无源汇可行流

    题目: 无源汇可行流例题 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1314 题解: 证明什么的就算了,下面给出一种建图方式 ...

  5. HDU 4940 Destroy Transportation system(无源汇上下界网络流)

    Problem Description Tom is a commander, his task is destroying his enemy’s transportation system. Le ...

  6. hdu 4940 Destroy Transportation system( 无源汇上下界网络流的可行流推断 )

    题意:有n个点和m条有向边构成的网络.每条边有两个花费: d:毁坏这条边的花费 b:重建一条双向边的花费 寻找这样两个点集,使得点集s到点集t满足 毁坏全部S到T的路径的费用和 > 毁坏全部T到 ...

  7. ZOJ 2314 Reactor Cooling(无源汇上下界网络流)

    http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2314 题意: 给出每条边流量的上下界,问是否存在可行流,如果存在则输出. ...

  8. POJ 2396 Budget(有源汇上下界网络流)

    Description We are supposed to make a budget proposal for this multi-site competition. The budget pr ...

  9. zoj 2314 Reactor Cooling (无源汇上下界可行流)

    Reactor Coolinghttp://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1314 Time Limit: 5 Seconds ...

随机推荐

  1. DP入门

    数塔HDU2084 #include <iostream> #include <algorithm> #include <cstdio> #include < ...

  2. typedef和define具体的详细区别

    1) #define是预处理指令,在编译预处理时进行简单的替换,不作正确性检查,不关含义是否正确照样带入,只有在编译已被展开的源程序时才会发现可能的错误并报错.例如: #define PI 3.141 ...

  3. 蓝桥杯 0/1背包问题 (java)

      今天第一次接触了0/1背包问题,总结一下,方便以后修改.不对的地方还请大家不啬赐教! 上一个蓝桥杯的例题: 数据规模和约定 代码: import java.util.Scanner; public ...

  4. POJ 1502 MPI Maelstrom(模板题——Floyd算法)

    题目: BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distri ...

  5. ZOJ 1203 Swordfish

    题目: There exists a world within our world A world beneath what we call cyberspace. A world protected ...

  6. JavaScript八张思维导图—编程风格

    JS基本概念 JS操作符 JS基本语句 JS数组用法 Date用法 JS字符串用法 JS编程风格 JS编程实践 不知不觉做前端已经五年多了,无论是从最初的jQuery还是现在火热的Angular,Vu ...

  7. Spring MVC集成Swagger

    什么是Swagger? 大部分 Web 应用程序都支持 RESTful API,但不同于 SOAP API——REST API 依赖于 HTTP 方法,缺少与 Web 服务描述语言(Web Servi ...

  8. Oracle_子查询

    Oracle_子查询 子查询   --如何查得所有比"CLARK"工资高的员工的信息 select ename, sal from emp where ename = 'CLARK ...

  9. 虚拟主机导入MySQL出现Unknown character set: ‘utf8mb4’

    http://www.lmlblog.com/14.html 前几天进行网站搬家,MySQL导入数据的时候,出现以下错误(没有定义的编码集utf8mb4): SQL 查询: ; MySQL 返回:文档 ...

  10. asm文件开头的assume意义

    body, table{font-family: Consolas; font-size: 13.5pt} table{border-collapse: collapse; border: solid ...