简单的拓扑图dp..

A(i, j), B(i, j) 表示从点 i 长度为 j 的两种路径是否存在. 用bitset就行了

时间复杂度O(m)

----------------------------------------------------------------

#include<bits/stdc++.h>
 
#define clr(x, c) memset(x, c, sizeof(x))
#define rep(i, n) for(int i = 0; i < n; ++i)
#define foreach(i, x) for(__typeof(x.begin()) i = x.begin(); i != x.end(); i++)
 
using namespace std;
 
const int maxn = 109;
 
struct edge {
int to, a, b;
};
 
vector<edge> G[maxn];
queue<int> Q;
bitset<10009> A[maxn], B[maxn];
int n, inD[maxn] = {0};
 
int main() {
freopen("test.in", "r", stdin);
int m;
cin >> n >> m;
while(m--) {
int u, v, a, b;
scanf("%d%d%d%d", &u, &v, &a, &b); u--; v--;
G[u].push_back( (edge) {v, a, b} );
inD[v]++;
}
rep(i, n) {
   if(!inD[i]) Q.push(i);
A[i].reset(); B[i].reset();
}
A[0][0] = 1; B[0][0] = 1;
while(!Q.empty()) {
int x = Q.front(); Q.pop();
foreach(it, G[x]) {
A[it->to] |= A[x] << it->a;
B[it->to] |= B[x] << it->b;
if(!--inD[it->to]) Q.push(it->to);
}
}
int ans = -1;
rep(i, 10001) if(A[n - 1][i] && B[n - 1][i]) {
   ans = i;
   break;
}
if(!~ans)
   puts("IMPOSSIBLE");
else
   printf("%d\n", ans);
return 0;
}

----------------------------------------------------------------

3890: [Usaco2015 Jan]Meeting Time

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 95  Solved: 65
[Submit][Status][Discuss]

Description

Bessie and her sister Elsie want to travel from the barn to their favorite field, such that they leave at exactly the same time from the barn, and also arrive at exactly the same time at their favorite field. The farm is a collection of N fields (1 <= N <= 100) numbered 1..N, where field 1 contains the barn and field N is the favorite field. The farm is built on the side of a hill, with field X being higher in elevation than field Y if X < Y. An assortment of M paths connect pairs of fields. However, since each path is rather steep, it can only be followed in a downhill direction. For example, a path connecting field 5 with field 8 could be followed in the 5 -> 8 direction but not the other way, since this would be uphill. Each pair of fields is connected by at most one path, so M <= N(N-1)/2. It might take Bessie and Elsie different amounts of time to follow a path; for example, Bessie might take 10 units of time, and Elsie 20. Moreover, Bessie and Elsie only consume time when traveling on paths between fields -- since they are in a hurry, they always travel through a field in essentially zero time, never waiting around anywhere. Please help determine the shortest amount of time Bessie and Elsie must take in order to reach their favorite field at exactly the same moment.
给出一个n个点m条边的有向无环图,每条边两个边权。 
n<=100,没有重边。 
然后要求两条长度相同且尽量短的路径, 
路径1采用第一种边权,路径2采用第二种边权。 
没有则输出”IMPOSSIBLE”

Input

The first input line contains N and M, separated by a space. Each of the following M lines describes a path using four integers A B C D, where A and B (with A < B) are the fields connected by the path, C is the time required for Bessie to follow the path, and D is the time required for Elsie to follow the path. Both C and D are in the range 1..100.

Output

A single integer, giving the minimum time required for Bessie and Elsie to travel to their favorite field and arrive at the same moment. If this is impossible, or if there is no way for Bessie or Elsie to reach the favorite field at all, output the word IMPOSSIBLE on a single line.

Sample Input

3 3
1 3 1 2
1 2 1 2
2 3 1 2

Sample Output

2

SOLUTION NOTES:

Bessie is twice as fast as Elsie on each path, but if Bessie takes the
path 1->2->3 and Elsie takes the path 1->3 they will arrive at the
same time.

HINT

Source

3890: [Usaco2015 Jan]Meeting Time( dp )的更多相关文章

  1. BZOJ 3890 [Usaco2015 Jan]Meeting Time:拓扑图dp

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=3890 题意: 给你一个有向图,n个点(n <= 100),m条边. 且所有的边都是从 ...

  2. bzoj3890 [Usaco2015 Jan]Meeting Time

    Description Bessie and her sister Elsie want to travel from the barn to their favorite field, such t ...

  3. BZOJ3887 [Usaco2015 Jan] Grass Cownoisseur 【tarjan】【DP】*

    BZOJ3887 [Usaco2015 Jan] Grass Cownoisseur Description In an effort to better manage the grazing pat ...

  4. BZOJ_3887_[Usaco2015 Jan]Grass Cownoisseur_强连通分量+拓扑排序+DP

    BZOJ_3887_[Usaco2015 Jan]Grass Cownoisseur_强连通分量+拓扑排序+DP Description In an effort to better manage t ...

  5. [bzoj3887][Usaco2015 Jan]Grass Cownoisseur_trajan_拓扑排序_拓扑序dp

    [Usaco2015 Jan]Grass Cownoisseur 题目大意:给一个有向图,然后选一条路径起点终点都为1的路径出来,有一次机会可以沿某条边逆方向走,问最多有多少个点可以被经过?(一个点在 ...

  6. [补档][Usaco2015 Jan]Grass Cownoisseur

    [Usaco2015 Jan]Grass Cownoisseur 题目 给一个有向图,然后选一条路径起点终点都为1的路径出来,有一次机会可以沿某条边逆方向走,问最多有多少个点可以被经过? (一个点在路 ...

  7. bzoj3887: [Usaco2015 Jan]Grass Cownoisseur

    题意: 给一个有向图,然后选一条路径起点终点都为1的路径出来,有一次机会可以沿某条边逆方向走,问最多有多少个点可以被经过?(一个点在路径中无论出现多少正整数次对答案的贡献均为1) =>有向图我们 ...

  8. 【bzoj3886】[Usaco2015 Jan]Moovie Mooving 状态压缩dp+二分

    题目描述 Bessie is out at the movies. Being mischievous as always, she has decided to hide from Farmer J ...

  9. BZOJ 1677: [Usaco2005 Jan]Sumsets 求和( dp )

    完全背包.. --------------------------------------------------------------------------------------- #incl ...

随机推荐

  1. HTML解决div里面img的缝隙问题

    图片IMG与容器下边界之间有空隙怎么办?这里介绍3中简单的解决方法. 第一,给图片img标签display:block. img{display:block} 第二,定义容器里的字体大小为0. div ...

  2. windows phone:使用sqlite-net

    继上篇文章后,这里简单介绍下sqlite-net的使用(示例不为作者所写,摘自于:https://github.com/peterhuene/sqlite-net) Please consult th ...

  3. UIColor深入研究(CGColor,CIColor)

    由于跟人比较喜欢研究关于图层与动画方面的技术,正打算看看别人写的好东西,就遇到了好几个问题, 第一:UIClor类方法的使用 就是关于UIColor的使用,记得之前开发中我们使用的都是UIColor后 ...

  4. Parrot源代码分析之海贼王

    我们的目的是找到speedup-example在使用Parrot加速的原因,假设仅仅说它源于Context Switch的降低,有点简单了,它究竟为什么降低了?除了Context Switch外是否还 ...

  5. stm32之GPIO库函数开发

    关于GPIO库函数的重点函数:P122 GPIO_Init() :根据GPIO_InitStruct中指定的参数初始化外设GPIOx寄存器: GPIO_ReadInputDataBit():读取指定端 ...

  6. php5.5 + apache2.4 安装配置

    php5.5 做了大量的更新,在与apache搭配的时候如何选择也很有讲究,这里我们以64位 php5.6 和 Apache2.4为例介绍如何配置. 工具/原料 Win7/8 64位 php5.5.6 ...

  7. tomcat环境变量的配置(网上摘,全部验证通过)

    tomcat环境变量的配置   1.===> 进入bin目录下,双击startup.bat看是否报错.一般肯定会报. 2.===> 右键我的电脑===>高级===>环境变量   ...

  8. USACO Cow Pedigrees 【Dp】

    一道经典Dp. 定义dp[i][j] 表示由i个节点,j 层高度的累计方法数 状态转移方程为: 用i个点组成深度最多为j的二叉树的方法树等于组成左子树的方法数 乘于组成右子树的方法数再累计. & ...

  9. POJ 3675 Telescope 简单多边形和圆的面积交

    这道题得控制好精度,不然会贡献WA  QAQ 还是那个规则: int sgn(double x){ if(x > eps) return 1; else if(x < - eps) ret ...

  10. C#实现时间戳转化

    /// <summary> /// 时间戳转为C#格式时间 /// </summary> /// <param name=”timeStamp”></para ...