River Problem

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 721    Accepted Submission(s): 282

Problem Description
The River of Bitland is now heavily polluted. To solve this problem, the King of Bitland decides to use some kinds of chemicals to make the river clean again.

The structure of the river contains n nodes and exactly n-1 edges between those nodes. It's just the same as all the rivers in this world: The edges are all unidirectional to represent water flows. There are source points, from which the water flows, and there is exactly one sink node, at which all parts of the river meet together and run into the sea. The water always flows from sources to sink, so from any nodes we can find a directed path that leads to the sink node. Note that the sink node is always labeled 1.

As you can see, some parts of the river are polluted, and we set a weight Wi for each edge to show how heavily polluted this edge is. We have m kinds of chemicals to clean the river. The i-th chemical can decrease the weight for all edges in the path from Ui to Vi by exactly 1. Moreover, we can use this kind of chemical for Li times, the cost for each time is Ci. Note that you can still use the chemical even if the weight of edges are 0, but the weight of that edge will not decrease this time.

When the weight of all edges are 0, the river is cleaned, please help us to clean the river with the least cost.

 
Input
The first line of the input is an integer T representing the number of test cases. The following T blocks each represents a test case.

The first line of each block contains a number n (2<=n<=150) representing the number of nodes. The following n-1 lines each contains 3 numbers U, V, and W, means there is a directed edge from U to V, and the pollution weight of this edge is W. (1<=U,V<=n, 0<=W<=20)

Then follows an number m (1<=m<=2000), representing the number of chemical kinds. The following m lines each contains 4 numbers Ui, Vi, Li and Ci (1<=Ui,Vi<=n, 1<=Li<=20, 1<=Ci<=1000), describing a kind of chemical, as described above. It is guaranteed that from Ui we can always find a directed path to Vi.

 
Output
First output "Case #k: ", where k is the case numbers, then follows a number indicating the least cost you are required to calculate, if the answer does not exist, output "-1" instead.
 
Sample Input
2
3
2 1 2
3 1 1
1
3 1 2 2
3
2 1 2
3 1 1
2
3 1 2 2
2 1 2 1
 
Sample Output
Case #1: -1
Case #2: 4
 
Author
Thost & Kennethsnow
 

Noi2008 志愿者招募 一样 就是相邻的节点  不是连续的天数了 而是建立了一个图

用dfs走一遍  建图就好了

公式不用推  看懂 那个题想一下就好了

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _ ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1e5 + , INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;
int n, m, s, t;
int head[maxn], d[maxn], vis[maxn], nex[maxn], f[maxn], p[maxn], cnt, head1[maxn], nex1[maxn];
int xu[maxn], flow, value, ans; struct edge
{
int u, v, c;
}Edge[maxn << ]; void addedge(int u, int v, int c)
{
Edge[ans].u = u;
Edge[ans].v = v;
Edge[ans].c = c;
nex1[ans] = head1[u];
head1[u] = ans++;
}; struct node
{
int u, v, w, c;
}Node[maxn << ]; void add_(int u, int v, int w, int c)
{
Node[cnt].u = u;
Node[cnt].v = v;
Node[cnt].w = w;
Node[cnt].c = c;
nex[cnt] = head[u];
head[u] = cnt++;
} void add(int u, int v, int w, int c)
{
add_(u, v, w, c);
add_(v, u, -w, );
} int spfa()
{
for(int i = ; i < maxn; i ++) d[i] = INF;
deque<int> Q;
mem(vis, );
mem(p, -);
Q.push_front(s);
d[s] = ;
p[s] = , f[s] = INF;
while(!Q.empty())
{
int u = Q.front(); Q.pop_front();
vis[u] = ;
for(int i = head[u];i != -; i = nex[i])
{
int v = Node[i].v;
if(Node[i].c)
{
if(d[v] > d[u] + Node[i].w)
{
d[v] = d[u] + Node[i].w;
p[v] = i;
f[v] = min(f[u], Node[i].c);
if(!vis[v])
{
// cout << v << endl;
if(Q.empty()) Q.push_front(v);
else
{
if(d[v] < d[Q.front()]) Q.push_front(v);
else Q.push_back(v);
}
vis[v] = ;
}
}
}
}
}
if(p[t] == -) return ;
flow += f[t], value += f[t] * d[t];
// cout << value << endl;
for(int i = t; i != s; i = Node[p[i]].u)
{
Node[p[i]].c -= f[t];
Node[p[i] ^ ].c += f[t];
}
return ;
} void max_flow()
{
flow = value = ;
while(spfa());
}
int sum_flow; void init()
{
mem(head, -);
mem(head1, -);
Edge[].c = ;
cnt = sum_flow = ;
ans = ;
} void dfs(int u, int pre_sum)
{
int sum = ;
for(int i = head1[u]; i != -; i = nex1[i])
{
int v = Edge[i].v;
add(u, v, , INF);
dfs(v, Edge[i].c);
sum += Edge[i].c; //要减去当前子节点的所有父节点的公式
}
int tmp = pre_sum - sum;
if(tmp > ) add(s, u, , tmp), sum_flow += tmp;
else add(u, t, , -tmp); } int id[maxn]; int main()
{
int T, kase = ;
int u, v, w, c;
rd(T);
while(T--)
{
init();
rd(n);
s = , t = n + ;
rap(i, , n - )
{
rd(u), rd(v), rd(w);
addedge(v, u, w); //反向建图 想一下是下一个公式减去上一个公式 即子结点减去父结点
}
addedge(t, , );
rd(m);
rap(i, , m)
{
rd(u), rd(v), rd(c), rd(w);
add(u, v, w, c);
}
dfs(, );
max_flow();
printf("Case #%d: ", ++kase);
if(sum_flow == flow)
cout << value << endl;
else
cout << - << endl; } return ;
}

River Problem

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 721    Accepted Submission(s): 282

Problem Description
The River of Bitland is now heavily polluted. To solve this problem, the King of Bitland decides to use some kinds of chemicals to make the river clean again.

The structure of the river contains n nodes and exactly n-1 edges between those nodes. It's just the same as all the rivers in this world: The edges are all unidirectional to represent water flows. There are source points, from which the water flows, and there is exactly one sink node, at which all parts of the river meet together and run into the sea. The water always flows from sources to sink, so from any nodes we can find a directed path that leads to the sink node. Note that the sink node is always labeled 1.

As you can see, some parts of the river are polluted, and we set a weight Wi for each edge to show how heavily polluted this edge is. We have m kinds of chemicals to clean the river. The i-th chemical can decrease the weight for all edges in the path from Ui to Vi by exactly 1. Moreover, we can use this kind of chemical for Li times, the cost for each time is Ci. Note that you can still use the chemical even if the weight of edges are 0, but the weight of that edge will not decrease this time.

When the weight of all edges are 0, the river is cleaned, please help us to clean the river with the least cost.

 
Input
The first line of the input is an integer T representing the number of test cases. The following T blocks each represents a test case.

The first line of each block contains a number n (2<=n<=150) representing the number of nodes. The following n-1 lines each contains 3 numbers U, V, and W, means there is a directed edge from U to V, and the pollution weight of this edge is W. (1<=U,V<=n, 0<=W<=20)

Then follows an number m (1<=m<=2000), representing the number of chemical kinds. The following m lines each contains 4 numbers Ui, Vi, Li and Ci (1<=Ui,Vi<=n, 1<=Li<=20, 1<=Ci<=1000), describing a kind of chemical, as described above. It is guaranteed that from Ui we can always find a directed path to Vi.

 
Output
First output "Case #k: ", where k is the case numbers, then follows a number indicating the least cost you are required to calculate, if the answer does not exist, output "-1" instead.
 
Sample Input
2
3
2 1 2
3 1 1
1
3 1 2 2
3
2 1 2
3 1 1
2
3 1 2 2
2 1 2 1
 
Sample Output
Case #1: -1
Case #2: 4
 
Author
Thost & Kennethsnow
 

River Problem HDU - 3947(公式建边)的更多相关文章

  1. HDU 3947 River Problem

    River Problem Time Limit: 2000ms Memory Limit: 65536KB This problem will be judged on HDU. Original ...

  2. Flow Problem HDU - 3549

    Flow Problem HDU - 3549 Network flow is a well-known difficult problem for ACMers. Given a graph, yo ...

  3. D - Ugly Problem HDU - 5920

    D - Ugly Problem HDU - 5920 Everyone hates ugly problems. You are given a positive integer. You must ...

  4. Prime Ring Problem HDU - 1016 (dfs)

    Prime Ring Problem HDU - 1016 A ring is compose of n circles as shown in diagram. Put natural number ...

  5. 志愿者招募 HYSBZ - 1061(公式建图费用流)

    转自神犇:https://www.cnblogs.com/jianglangcaijin/p/3799759.html 题意:申奥成功后,布布经过不懈努力,终于 成为奥组委下属公司人力资源部门的主管. ...

  6. HDU 3947 Assign the task

    http://acm.hdu.edu.cn/showproblem.php?pid=3974 Problem Description There is a company that has N emp ...

  7. (线段树 区间查询)The Water Problem -- hdu -- 5443 (2015 ACM/ICPC Asia Regional Changchun Online)

    链接: http://acm.hdu.edu.cn/showproblem.php?pid=5443 The Water Problem Time Limit: 1500/1000 MS (Java/ ...

  8. 差分约束系统+(矩阵)思维(H - THE MATRIX PROBLEM HDU - 3666 )

    题目链接:https://cn.vjudge.net/contest/276233#problem/H 题目大意:对于给定的矩阵  每一行除以ai  每一列除以bi 之后 数组的所有元素都还在那个L- ...

  9. HDU 4522 (恶心建图)

    湫湫系列故事——过年回家 Time Limit: 500/200 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total ...

随机推荐

  1. 通过 Systemd Journal 收集日志

    随着 systemd 成了主流的 init 系统,systemd 的功能也在不断的增加,比如对系统日志的管理.Systemd 设计的日志系统好处多多,这里笔者就不再赘述了,本文笔者主要介绍 syste ...

  2. configure: error: cannot guess build type; you must specify one解决方法

    原文地址:https://blog.csdn.net/hebbely/article/details/53993141 1.configure: error: cannot guess build t ...

  3. Centos7 下SVN迁移

    SVN迁移需要做如下操作: 1. 将原来的Repository导出 . #svnadmin dump 原有repos的目录路径 > dumpfile (不同服务器安装目录不同,根据具体情况调整) ...

  4. hdu6249 区间动态规划

    题目链接 题意:给出一些区间,求选k个区间能覆盖的最多点的数量 思路:定义dp[i][j]为前i个点取j个区间的最大值.dp[i][j]可以转移到dp[i+1][j+1]和以i+1为起点的区间终点 具 ...

  5. Day5 Pyhton基础之编码与解码(四)

    1.编码与解码 1.1现在常用的编码类型

  6. java得到日期相减的天数

    /** * <li>功能描述:时间相减得到天数 * @param beginDateStr * @param endDateStr * @return * long * @author A ...

  7. 使用git将本地项目推送到码云私有仓库

    https://blog.csdn.net/qq_33876553/article/details/80111946 2018年04月27日 19:53:33 桥路丶 阅读数:2958 前言 之前博主 ...

  8. CentOS 6.5 手动rpm包安装gcc、g++

    摘自:https://blog.csdn.net/lichen_net/article/details/70211204 mount CentOS的安装光盘,然后先后安装 rpm -ivh ppl-0 ...

  9. [2017BUAA软工助教]个人项目测试结果

    个人项目测试结果 标签(空格分隔): 未分类 9.29第一次测试结果 注:点击表头内相应项目可针对该项目进行排序 -c测试结果 INDEX NumberID -c 1 -c 5 -c 100 -c 5 ...

  10. 修改eclipce操作权限

    <dependencies> <dependency> <groupId>jdk.tools</groupId> <artifactId>j ...