hdu 4109(拓扑排序 关键路径)
来源:互联网 发布:php 抓取 天猫 商品 编辑:程序博客网 时间:2024/06/16 08:12
与hdu2647有相似之处,需要牢记。
题目:http://acm.hdu.edu.cn/showproblem.php?pid=4109
Instrction Arrangement
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1268 Accepted Submission(s): 525
Problem Description
Ali has taken the Computer Organization and Architecture course this term. He learned that there may be dependence between instructions, like WAR (write after read), WAW, RAW.
If the distance between two instructions is less than the Safe Distance, it will result in hazard, which may cause wrong result. So we need to design special circuit to eliminate hazard. However the most simple way to solve this problem is to add bubbles (useless operation), which means wasting time to ensure that the distance between two instructions is not smaller than the Safe Distance.
The definition of the distance between two instructions is the difference between their beginning times.
Now we have many instructions, and we know the dependent relations and Safe Distances between instructions. We also have a very strong CPU with infinite number of cores, so you can run as many instructions as you want simultaneity, and the CPU is so fast that it just cost 1ns to finish any instruction.
Your job is to rearrange the instructions so that the CPU can finish all the instructions using minimum time.
If the distance between two instructions is less than the Safe Distance, it will result in hazard, which may cause wrong result. So we need to design special circuit to eliminate hazard. However the most simple way to solve this problem is to add bubbles (useless operation), which means wasting time to ensure that the distance between two instructions is not smaller than the Safe Distance.
The definition of the distance between two instructions is the difference between their beginning times.
Now we have many instructions, and we know the dependent relations and Safe Distances between instructions. We also have a very strong CPU with infinite number of cores, so you can run as many instructions as you want simultaneity, and the CPU is so fast that it just cost 1ns to finish any instruction.
Your job is to rearrange the instructions so that the CPU can finish all the instructions using minimum time.
Input
The input consists several testcases.
The first line has two integers N, M (N <= 1000, M <= 10000), means that there are N instructions and M dependent relations.
The following M lines, each contains three integers X, Y , Z, means the Safe Distance between X and Y is Z, and Y should run after X. The instructions are numbered from 0 to N - 1.
The first line has two integers N, M (N <= 1000, M <= 10000), means that there are N instructions and M dependent relations.
The following M lines, each contains three integers X, Y , Z, means the Safe Distance between X and Y is Z, and Y should run after X. The instructions are numbered from 0 to N - 1.
Output
Print one integer, the minimum time the CPU needs to run.
Sample Input
5 21 2 13 4 1
Sample Output
2HintIn the 1st ns, instruction 0, 1 and 3 are executed;In the 2nd ns, instruction 2 and 4 are executed.So the answer should be 2.
由入度为0的点开始,按照层数递进,算出每个任务点完成所需最少时间,最后一一比较即可。
#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdlib>#include <string>#include <vector>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>using namespace std;struct node{ int y,t;}edge[10005];vector<node> v[1005];int n,m;int in[1005],ti[1005];void solve(){ queue<int> q; for (int i=0; i<n; i++) { if (in[i]==0) { q.push(i); ti[i]=1; } } while (!q.empty()) { int x=q.front();q.pop(); for (int i=0; i<v[x].size(); i++) { int y=v[x][i].y; ti[y]=max(v[x][i].t+ti[x],ti[y]); in[y]--; if (in[y]==0) { q.push(y); } } } int ans=0; for (int i=0; i<n; i++) { ans=max(ans,ti[i]); } cout<<ans<<endl;}int main(){ int x; while (cin>>n>>m) { memset(in, 0, sizeof(in)); memset(ti, 0, sizeof(ti)); for (int i=0; i<n; i++) { v[i].clear(); } for (int i=0; i<m; i++) { cin>>x>>edge[i].y>>edge[i].t; v[x].push_back(edge[i]); in[edge[i].y]++; } solve(); } return 0;}
0 0
- hdu 4109 拓扑排序 关键路径
- hdu 4109(拓扑排序 关键路径)
- hdu 4109 Instrction Arrangement 拓扑排序 关键路径
- hdu 4109 Instrction Arrangement 拓扑排序 关键路径
- HDU 4109 Instrction Arrangement拓扑排序 关键路径模板
- HDU 4109 拓扑排序(最短路思想)关键路径
- 拓扑排序关键路径
- 拓扑排序,关键路径
- 拓扑排序&关键路径
- 拓扑排序--关键路径
- 关键路径 + 拓扑排序
- 拓扑排序和关键路径
- 拓扑排序与关键路径
- 拓扑排序--关键路径实现
- 拓扑排序和关键路径
- 拓扑排序和关键路径
- 拓扑排序之关键路径
- 拓扑排序和关键路径
- 求大于200的最小质数,java
- 21-IO流-08-IO流(字符流-练习-复制文本文件_1)
- Codeforces Round #315 (Div. 2) A. Music
- nfs挂载和取消挂载命令
- 《Spring 2.0技术手册》 读书笔记七-Spring的DAO框架(3)-JDBC事务管理
- hdu 4109(拓扑排序 关键路径)
- NSString 查找指定字符串出现的次数
- 文章标题
- SDUTOJ懒虫小鑫
- Leetcode#5||Longest Palindromic Substring
- 21-IO流-09-IO流(字符流-练习-复制文本文件_2)
- 图片
- CodeForces - 344D Alternating Current (模拟题)
- Linux C基于Socket的多线程扫描程序