hust1608 Dating With Girls hust校赛 BFS求最长路
http://acm.hust.edu.cn/problem.php?id=1608
Dating With Girls
Time Limit: 1 Sec Memory Limit: 128 MBSubmissions: 146 Solved: 22
Description
If you have a date with a pretty girl in a hurry, you can ignore what I will say next.
Hellis is a little bad guy in Rural Small Technical College. And the most important fact is that he is especially fond of glaring at pretty girls! And as we all know, there are some girls he favors in the same school. So there comes the trouble. On one terrible day, it should rains. The girls all are being different places. They all phone him and ask the boy to deliver them umbrellas at the same time. However, the cute boy faces the embarrassing condition. You can have a simple Understanding that each girl has a special relationship with our sunny boy. If they know the boy sends the umbrella to anyone of them, the others may go mad and that is not the boy expects. If that happens, Of course you will never see that he is completing again. But the fact is some girls are so beautiful and he would like to give them help. The trouble is this guy wants to meet more beautiful girls before he arrives at anyone who phones him for help. It is just a disaster, he thinks. Eventually, he makes the choice. He will just send an umbrella to only one girl who is most important to him, and he can not be seen by other girls who phones him for help. If that happens, I can only say hehe. He must pass these girls. In the process, he can send beautiful girls their umbrellas! In that case, he can create a chance to communicate with them and just win their favorable impression. It is such a good idea!
There are n different places in our problem. There are m different undirectional edges. And there is no loop. The bad guy starts at 1 node, and other 2 to n node stand different girls. Each girl is standing at different nodes, too. The i-th girl has wi. When Hellis meets a girl, he can get |wi| point, and he wants to get max sum of point he gets.(wi<0 mean the i-th girl has phoned him for help
Input
First line, two integers, n ,m (1<n<100, 1<m<5000)
2th to (m+1)th per line ,ai, bi (0<ai<=n, 0<bi<=n) means one road from ai to bi.
(m+2)th to (n+m)th per line, wi (0<|wi|<=100, 2<=i<=n)
Output
If the guy can meet the girl they chose, output line print a single integer ans — the max sum of point he gets.
Else print “What is a fucking day!”
Sample Input
3 3 1 2 1 3 2 3 -30 10
Sample Output
30
HINT
Source
The 7th(2012) ACM Programming Contest of HUST
Problem Setter: Senlan Yao
/*
题意:
一个图上有n个点，男孩在1号点，女孩在2至n个点，每个女孩有一个值wi，问男孩到达最小wi值（wi保证最小的小于0）
的那个女孩且中间不经过任何wi<0的女孩并获得最多的值为多少？，男孩每经过一个女孩他将获得|wi|的值；
思路：以1好点为起点，求一个终点为wi且（wi<0，中间不经过任何wi<0的点）的最长路，输出wi<0中wi最小的那个所得的值；
*/
#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
using namespace std;
int vis[105],map[105][105],val[105],mmax[105],n;
void BFS()
{
int i;
memset(vis,0,sizeof(vis));
for(i=1;i<=n;i++)
mmax[i]=-1000;
queue<int>que;
mmax[1]=0;
que.push(1);
vis[1]=1;
while(!que.empty())
{
int temp=que.front();
que.pop();
for(i=1;i<=n;i++)
{
if(map[temp][i])/*如果temp到i有路可走 那么先不管是否通过其它路径走过 先看通过现有路径能不能使该点更长 如果可以那么就更新*/
{
if(val[i]>0)//如果val[i]>0 即i点对应的值大于0直接加
{
if(mmax[i]<mmax[temp]+val[i])
mmax[i]=mmax[temp]+val[i];
}
else//如果val[i]<0 即i点对应的值<于0加其相反数
{
if(mmax[i]<mmax[temp]-val[i])
mmax[i]=mmax[temp]-val[i];
}
if(!vis[i]&&val[i]>0)//如果该点已经走过就没有必要再入队了 另外根据题意要求最后一个点小于0 所以不能把小于0的点入队
{
que.push(i);
vis[i]=1;
}
}
}
vis[temp]=0;
}
}
int main()
{
int i,m,x,y,min,flag;
while(scanf("%d %d",&n,&m)!=EOF)
{
memset(map,0,sizeof(map));
while(m--)
{
scanf("%d %d",&x,&y);
map[x][y]=1;
}
min=10000;
for(i=2;i<=n;i++)
{
scanf("%d",&val[i]);
if(min>val[i]) {min=val[i];flag=i;}
}
BFS();
if(mmax[flag]>=0)
printf("%d\n",mmax[flag]);
else printf("What is a fucking day!\n");
}
return 0;
}