Reactor Cooling
- Time Limit: 5 Seconds
- Memory Limit: 32768 KB
- Special Judge
The terrorist group leaded by a well known international terrorist Ben Bladen is buliding a nuclear reactor to produce plutonium for the nuclear bomb they are planning to create. Being the wicked computer genius of this group, you are responsible for developing the cooling system for the reactor.
The cooling system of the reactor consists of the number of pipes that special cooling liquid flows by. Pipes are connected at special points, called nodes, each pipe has the starting node and the end point. The liquid must flow by the pipe from its start point to its end point and not in the opposite direction.
Let the nodes be numbered from 1 to N. The cooling system must be designed so that the liquid is circulating by the pipes and the amount of the liquid coming to each node (in the unit of time) is equal to the amount of liquid leaving the node. That is, if we designate the amount of liquid going by the pipe from i-th node to j-th as fij, (put fij = 0 if there is no pipe from node i to node j), for each i the following condition must hold:
fi,1+fi,2+...+fi,N = f1,i+f2,i+...+fN,i
Each pipe has some finite capacity, therefore for each i and j connected by the pipe must be fij <= cij where cij is the capacity of the pipe. To provide sufficient cooling, the amount of the liquid flowing by the pipe going from i-th to j-th nodes must be at least lij, thus it must be fij >= lij.
Given cij and lij for all pipes, find the amount fij, satisfying the conditions specified above.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
Input
The first line of the input file contains the number N (1 <= N <= 200) - the number of nodes and and M - the number of pipes. The following M lines contain four integer number each - i, j, lij and cij each. There is at most one pipe connecting any two nodes and 0 <= lij <= cij <= 10^5 for all pipes. No pipe connects a node to itself. If there is a pipe from i-th node to j-th, there is no pipe from j-th node to i-th.
Output
On the first line of the output file print YES if there is the way to carry out reactor cooling and NO if there is none. In the first case M integers must follow, k-th number being the amount of liquid flowing by the k-th pipe. Pipes are numbered as they are given in the input file.
Sample Input
2
4 6
1 2 1 2
2 3 1 2
3 4 1 2
4 1 1 2
1 3 1 2
4 2 1 2
4 6
1 2 1 3
2 3 1 3
3 4 1 3
4 1 1 3
1 3 1 3
4 2 1 3Sample Output
NO
YES
1
2
3
2
1
1链接
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2314
题意
给n个点，及m根pipe，每根pipe用来流躺液体的，单向的，每时每刻每根pipe流进来的物质要等于流出去的物质，要使得m条pipe组成一个循环体，里面流躺物质。并且满足每根pipe一定的流量限制，范围为[Li,Ri].即要满足每时刻流进来的不能超过Ri(最大流问题)，同时最小不能低于Li。
题解
无源汇有上下界可行流。练习模板。
上界用ci表示，下界用bi表示。
下界是必须流满的，那么对于每一条边，去掉下界后，其自由流为ci– bi。
主要思想：每一个点流进来的流=流出去的流
对于每一个点i，令 Mi= sum(i点所有流进来的下界流)– sum(i点所有流出去的下界流)
如果Mi大于0，代表此点必须还要流出去Mi的自由流，那么我们从源点连一条Mi的边到该点。
如果Mi小于0，代表此点必须还要流进来Mi的自由流，那么我们从该点连一条Mi的边到汇点。
如果求S->T的最大流，看是否满流(S的相邻边都流满)。
满流则有解，否则无解。
- 转载自黄学长博客：传送门。
- 相关学习资料：有上下界的网络流学习笔记。
代码
| Status | Accepted |
|---|---|
| Time | 20ms |
| Memory | 4336kB |
| Length | 2992 |
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 210;
const int maxm = 200010;
const int inf = 0x3f3f3f3f;
int N, M;
struct Edge
{
int u, v, cap;
Edge() {}
Edge(int u, int v, int cap): u(u), v(v), cap(cap) {}
} es[maxm];
int R, S, T;
vector<int> tab[maxn]; // 边集
int dis[maxn];
int current[maxn];
int low[maxm], sum[maxn];
void addedge(int u, int v, int cap)
{
tab[u].push_back(R);
es[R++] = Edge(u, v, cap); // 正向边
tab[v].push_back(R);
es[R++] = Edge(v, u, 0); // 反向边容量为0
// 正向边下标通过异或就得到反向边下标, 2 ^ 1 == 3 ; 3 ^ 1 == 2
}
int BFS()
{
queue<int> q;
q.push(S);
memset(dis, 0x3f, sizeof(dis));
dis[S] = 0;
while (!q.empty())
{
int h = q.front();
q.pop();
for (int i = 0; i < tab[h].size(); i++)
{
Edge &e = es[tab[h][i]];
if (e.cap > 0 && dis[e.v] == 0x3f3f3f3f)
{
dis[e.v] = dis[h] + 1;
q.push(e.v);
}
}
}
return dis[T] < 0x3f3f3f3f; // 返回是否能够到达汇点
}
int dinic(int x, int maxflow)
{
if (x == T)
return maxflow;
// i = current[x] 当前弧优化
for (int i = current[x]; i < tab[x].size(); i++)
{
current[x] = i;
Edge &e = es[tab[x][i]];
if (dis[e.v] == dis[x] + 1 && e.cap > 0)
{
int flow = dinic(e.v, min(maxflow, e.cap));
if (flow)
{
e.cap -= flow; // 正向边流量降低
es[tab[x][i] ^ 1].cap += flow; // 反向边流量增加
return flow;
}
}
}
return 0; // 找不到增广路 退出
}
int DINIC()
{
int ans = 0;
while (BFS()) // 建立分层图
{
int flow;
memset(current, 0, sizeof(current)); // BFS后应当清空当前弧数组
while (flow = dinic(S, 0x3f3f3f3f)) // 一次BFS可以进行多次增广
ans += flow;
}
return ans;
}
int main()
{
int Kase, u, v, cap;
scanf("%d", &Kase);
while (Kase--)
{
scanf("%d%d", &N, &M);
S = 0, T = N + 1, R = 0;
for (int i = 0; i <= T; ++i)
{
sum[i] = 0;
tab[i].clear();
}
for (int i = 0; i < M; ++i)
{
scanf("%d%d%d%d", &u, &v, &low[i], &cap);
addedge(u, v, cap - low[i]); // 初始流为流入-流出
sum[u] -= low[i], sum[v] += low[i]; // 记录点流入流出情况
}
int tot = 0;
for (int i = 1; i <= N; ++i)
{
if (sum[i] > 0) // 流入量大于流出量，需要S与i连一条边
{
addedge(S, i, sum[i]);
tot += sum[i];
}
else // 流出量大于流入量，需要i与T连一条边
addedge(i, T, -sum[i]);
}
int ans = DINIC();
if (ans == tot) // 最大流等于源点流出流量，有解
{
printf("YES\n");
int id = 1;
for (int i = 0; i < M; ++i)
{
printf("%d\n", low[i] + es[id].cap); // 下界加上附加流
id += 2; // 边的编号
}
}
else
printf("NO\n");
}
return 0;
}慢慢补坑。
The end.
2018-09-13 2018年9月13日 星期四