称号:生命是非常多的选择.现在给你一些选择(0~n-1),和其他选项后,分支数每一次选择,选择共求. 分析:dp,图论.假设一个状态也许是选择的数量0一个是,代表死亡,计数的路径数将达到所有死亡可以去除. 用一个状态数组记录到达每一个选择的路径数,它等于能到达它的前驱节点的路径加和. 稀疏图,使用邻接表储存.初始是节点0的路径条数为1.代表出生. 说明:没有给数据范围略坑啊,RE一次.WA一次.╮(╯▽╰)╭. #include <iostream> #include <cstdlib&…

 Paths through the Hourglass Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 10564 #include <stdio.h> #include <string.h> int n,s; ][][]; ][]; int d(int x,int y,int m) { *n-) ; int v=a[x][y…

题意:给出源点和边,边权为1,让你求从源点出发的最长路径,求出路径长度和最后地点,若有多组,输出具有最小编号的最后地点. #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> using namespace std; ; int n,s,tot; //n:点的个数,s:源点 int dist[maxn]; /…

题目大意:给定n条单向边,求图中任意两点的连通路径的数目.其中点是从0-输入中出现的最大的点. 可以用floyd-warshall算法或者dfs. for(int k = 0; k < n; k++) for(int i = 0; i < n; i ++) for(int j = 0; j < n; j++) dp[i][j] += dp[i][k] * dp[k][j]; 这里这样写是不会重复的,原因在于k循环时,可能产生重复情况的 点对应的d[k][j]或d[i][k]为0,因此一条…

从下往上DP,d(i, j, k)表示第(i, j)个格子走到底和为k的路径条数. 至于字典序最小,DP的时候记录一下路径就好. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int n, sum; ][]; ][][]; ][][][]; int main() { && n) { memset(d, , sizeof(d)); memset(…

How an Event Enters a Cocoa Application An event is a low-level record of a user action that is usually routed to the application in which the action occurred. A typical event in OS X originates when the user manipulates an input device attached to a…

In one embodiment, a source device detects a packet flow that meets criteria for multi-path forwarding, and forwards a probe packet on a primary path from the source device to a destination device, the probe packet carrying an indication to cause a p…

UVA - 10564 Paths through the Hourglass 题意: 要求从第一层走到最下面一层,只能往左下或右下走 问有多少条路径之和刚好等于S? 如果有的话,输出字典序最小的路径. f[i][j][k]从下往上到第i层第j个和为k的方案数 上下转移不一样,分开处理 没必要判断走出沙漏 打印方案倒着找下去行了,尽量往左走   沙茶的忘注释掉文件WA好多次   #include <iostream> #include <cstdio> #include <a…

题目传送门 /* 01背包(类):dp[i][j][k] 表示从(i, j)出发的和为k的方案数,那么cnt = sum (dp[1][i][s]) 状态转移方程:dp[i][j][k] = dp[i+1][j][k-c] + dp[i+1][j+1][k-c];(下半部分) 上半部分类似 因为要输出字典序最小的,打印路径时先考虑L */ /************************************************ * Author :Running_Time * Crea…

原题链接在这里:https://leetcode.com/problems/all-paths-from-source-lead-to-destination/ 题目: Given the edges of a directed graph, and two nodes source and destination of this graph, determine whether or not all paths starting from source eventually end at de…

Double Shortest PathsAlice and Bob are walking in an ancient maze with a lot of caves and one-way passages connectingthem. They want to go from cave 1 to cave n. All the passages are difficult to pass. Passages are toosmall for two people to walk thr…

本文出自   http://blog.csdn.net/shuangde800 题目传送门 题意: 给一个相上面的图.要求从第一层走到最下面一层,只能往左下或右下走,经过的数字之和为sum. 问有多少条路径之和刚好等于S? 如果有的话,输出字典序最小的路径. 思路: f[i][j][k] 代表从(i,j)点往下走到最后一层和为k的方案数 那么,显然可以得到状态转移: f[i][j][k] = f[i+1][left][k-val] + f[i+1][right][k-val],  val=(i,…

为了方便打印路径,考虑从下往上转移.dp[i][j][S]表示在i行j列总和为S的方案, dp[i][j][S] = dp[i+1][left][S-x]+dp[i+1][right][S-x] 方案O(2^2*n-1),结果要用long long保存. #include<bits/stdc++.h> using namespace std; typedef long long ll; ,maxs = ; ][maxn]; ll dp[maxn<<][maxn][maxs]; //…

传送门:https://vjudge.net/problem/UVA-10564 题目大意:给你一张形如沙漏一般的图,每一个格子有一个权值,问你有多少种方案可以从第一行走到最后一行,并且输出起点最靠前的方案,以及此方案的起点编号,起点相同则字典序最小. 题解: 很容易想到一个DP,dp[i][j][S]代表到第i层,第j列,从第一层到这里的路径和为S的方案数,最后只要查询最后一行的方案数即可了.但是这样很不好输出方案!我们可能需要从终点向上dfs,但是又无法保证起点编号最小以及字典序最小.但是我…

题意: 由0-9的数字组成一个形如沙漏的图形,要求从第一行开始沿左下或者右下到达最后一行,问有多少种不同的路径,使最后路径上的整数之和为给定的某个数. 分析: 简单计数dp,从最后一行开始,设dp[i][j][k]为从下往上已经走过i行,走到第j列,此时路径上的整数之和为k的路径种数.得到递归方程: dp[i][j][k] += dp[i-1][j][k-a[i][j]] +dp[i-1][j + 1][k-a[i][j]]; 最后dfs输出路径,注意多条路径时要求坐标字典序最小! 代码: #i…

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=988 题目大意: 独轮车的车轮被分为5个扇形,分别涂上一种不同的颜色,现在有一个人行驶在M*N的玩个平面上.每个格子的大小刚好为一个扇形.有些格子有障碍,骑车的人从S出发要到达T,途中,在任何一个格子的时候他要么骑到下一个格子,要么左转或者右转90度,初始他面朝北,并且绿色格子贴着地面,要求到…

题目描述: 1015. Jill's Tour Paths Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description Every year, Jill takes a bicycle tour between two villages. There are different routes she can take between these villages, but she does have an upper limit…

题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=48 Tree Summing  Background LISP was one of the earliest high-level programming languages and, with FORTRAN, is one of the olde…

说明:关于Uva的题目,可以在vjudge上做的,不用到Uva(那个极其慢的)网站去做. 最小瓶颈路:找u到v的一条路径满足最大边权值尽量小 先求最小生成树,然后u到v的路径在树上是唯一的,答案就是这条路径. Uva 534 Frogger Time Limit: 3000MS 64bit IO Format: %lld & %llu Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly h…

题目链接:https://uva.onlinejudge.org/external/100/10047.pdf 题目链接:http://vjudge.net/contest/132239#problem/B <训练指南>P308 没什么好说的,学习一下刘汝佳的风格. #include <bits/stdc++.h> using namespace std; #define INF 0x3f3f3f3f int R,C; + ; + ; char maze[maxr][maxc];…

 Always on the run Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 590 Appoint description:  System Crawler  (2015-08-26) Description   Screeching tires. Searching lights. Wailing sirens. Police cars…

大白图论第二题··· 题意:独轮车的轮子被均分成五块,每块一个颜色,每走过一个格子恰好转过一个颜色. 在一个迷宫中,只能向前走或者左转90度或右转90度(我曾天真的认为是向左走和向右走···),每个操作的时间是1s. 在起点轮子的绿色块着地,方向向北,要求到终点时同样是绿色块着地,方向不限,求最短时间,若走不到输出"destination not reachable".每个样例中间空一行(没错我又因为这个wa了···uva上竟然没有pe···[捂脸跑开]) 解法:BFS.visit数组…

 UVa 1380 A Scheduling Problem 题目: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=41557 思路:   给出一个任务调度树,单向边u->v表示u必须在v之前完成,双向边u-v表示无所谓方向. 题目给出定理,首先dfs求得忽略无向边后的最长链点数k,那么问题就是判断是否可以通过无向边定向从而使得最长链点数不超过k.用dp的判断. 设f[i]表示以i为根的子树中所有的边定向后最长链点数不超过…

[题意] 有n个绿洲, m条道路,每条路上有一个温度,和一个路程长度,从绿洲s到绿洲t,求一条道路的最高温度尽量小, 如果有多条, 选一条总路程最短的. InputInput consists of several test cases. Your program must process all of them.The first line contains two integers N and E (1 ≤ N ≤ 100; 1 ≤ E ≤ 10000) where N represents…

  Longest Paths  It is a well known fact that some people do not have their social abilities completely enabled. One example is the lack of talent for calculating distances and intervals of time. This causes some people to always choose the longest w…

uva 116 Unidirectional TSP Background Problems that require minimum paths through some domain appear in many different areas of computer science. For example, one of the constraints in VLSI routing problems is minimizing wire length. The Traveling Sa…

这道题要求我们求出图中的给定的两个节点(一个起点一个终点,但这是无向图)之间所有“路径中最大权值”的最小值,这无疑是动态规划. 我开始时想到根据起点和终点用动态规划直接求结果,但最终由于题中S过大,会超时. 超时的代码如下: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <string…

解题报告 求最长路. 用SPFA求最长路,初始化图为零,dis数组也为零 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #define inf 99999999 #define N 110 using namespace std; int mmap[N][N],dis[N],vis[N],n; void…

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…

Problem    UVA - 10917 - Walk Through the Forest Time Limit: 3000 mSec Problem Description Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To…