先处理前两个学长到达各个点所需要的最少时间,在计算前两个学长和最后一个学长救出所有学妹的最少时间. #include<stdio.h> #include<string.h> #include<vector> #include<algorithm> using namespace std; const int inf=1000000000; struct node{ int b,d; }; struct node2{ int dst,sta; node2(){…

bfs+状态压缩求出所有的状态,然后由于第一个节点需要特殊处理,可以右移一位剔除掉,也可以特判.然后采用集合的操作, #pragma comment(linker,"/STACK:1024000000,1024000000") #include <cstdio> #include <queue> using namespace std; #define inf 0x3f3f3f3f int n, m, cnt; int head[17], next[17 * 1…

HDU 4280 Island Transport(网络流,最大流) Description In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the islands relies on the ships. You have a transportation company there. Some route…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4640 思路:f[i][j]表示一个人状态i下走到j的最小花费,dp[i][j]表示i个人在状态j下的最下花费.首先我们可以一遍bfs求出f[i][j],然后通过f[i][j]得到dp[1][i],最后就是更新dp[i][j]了. #include<iostream> #include<cstdio> #include<cstring> #include<algorit…

转载请注明出处:http://blog.csdn.net/u012860063 题目链接:pid=4280">http://acm.hdu.edu.cn/showproblem.php?pid=4280 Problem Description In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the is…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4640 解题思路: 首先用一个简单的2^n*n的dp可以求出一个人访问一个给定状态的最小花费,因为这i个人是等价的,所以用dp[i][mask]表示i个人跑完mask这个状态的最小花费,所以首先枚举集合mask,对于dp[i][mask],枚举mask的子集v,dp[i][mask]可以由dp[1][v],dp[i-1][mask^v]转移过来,注意这里用来合并的集合是不能有重复的,这个类似背包……这…

Island Transport Time Limit: 10000ms Memory Limit: 65536KB This problem will be judged on HDU. Original ID: 4280 64-bit integer IO format: %I64d      Java class name: Main In the vast waters far far away, there are many islands. People are living on…

Island Transport Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 6217    Accepted Submission(s): 1965 Problem Description In the vast waters far far away, there are many islands. People are l…

HDU 4280:http://acm.hdu.edu.cn/showproblem.php?pid=4280 题意: 比较裸的最大流题目,就是这是个无向图,并且比较卡时间. 思路: 是这样的,由于是无向图,所以addedge 的反边容量直接设为原始流量.然后还可以优化搜索的方向,bfs可以从t到s跑,dfs可以从s到t跑,这样快. //#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000&qu…

Island Transport Description In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the islands relies on the ships. You have a transportation company there. Some routes are opened for p…