hdu 1317 XYZZY
XYZZY
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4099 Accepted Submission(s):
1136
Problem Description
It has recently been discovered how to run open-source
software on the Y-Crate gaming device. A number of enterprising designers have
developed Advent-style games for deployment on the Y-Crate. Your job is to test
a number of these designs to see which are winnable.
Each game consists of a set of up to 100 rooms. One of the rooms is the start and one of the rooms is the finish. Each room has an energy value between -100 and +100. One-way doorways interconnect pairs of rooms.
The player begins in the start room with 100 energy points. She may pass through any doorway that connects the room she is in to another room, thus entering the other room. The energy value of this room is added to the player's energy. This process continues until she wins by entering the finish room or dies by running out of energy (or quits in frustration). During her adventure the player may enter the same room several times, receiving its energy each time.
Each game consists of a set of up to 100 rooms. One of the rooms is the start and one of the rooms is the finish. Each room has an energy value between -100 and +100. One-way doorways interconnect pairs of rooms.
The player begins in the start room with 100 energy points. She may pass through any doorway that connects the room she is in to another room, thus entering the other room. The energy value of this room is added to the player's energy. This process continues until she wins by entering the finish room or dies by running out of energy (or quits in frustration). During her adventure the player may enter the same room several times, receiving its energy each time.
Input
The input consists of several test cases. Each test
case begins with n, the number of rooms. The rooms are numbered from 1 (the
start room) to n (the finish room). Input for the n rooms follows. The input for
each room consists of one or more lines containing:
the energy value for room i
the number of doorways leaving room i
a list of the rooms that are reachable by the doorways leaving room i
The start and finish rooms will always have enery level 0. A line containing -1 follows the last test case.
the energy value for room i
the number of doorways leaving room i
a list of the rooms that are reachable by the doorways leaving room i
The start and finish rooms will always have enery level 0. A line containing -1 follows the last test case.
Output
In one line for each case, output "winnable" if it is
possible for the player to win, otherwise output "hopeless".
Sample Input
5
0 1 2
-60 1 3
-60 1 4
20 1 5
0 0
5
0 1 2
20 1 3
-60 1 4
-60 1 5
0 0
5
0 1 2
21 1 3
-60 1 4
-60 1 5
0 0
5
0 1 2
20 2 1 3
-60 1 4
-60 1 5
0 0
-1
Sample Output
hopeless
hopeless
winnable
winnable
Source
Recommend
这道题是我目前为止遇到过的最难的最短路，要运用一种新的算法，复制一下网上大神的思路。
1、先用Floyd算法判断图的连通性。如果1与n是不连通的，输出hopeless。
2、用Bellman_Ford算法判断是否有正圈，如果某点有正圈，并且该点与第n点是连通的。就输出winnable。当然，没有正圈的情况下，可以到达也是可以的。然后就是如何找正圈的问题。Bellman_Ford算法可以判断有没有负圈。Bellman_Ford是解决最短路问题的，核心是松弛法。如果dist[v]<dist[u]+Map[u][v],则dist[v]=dist[u]+Map[u][v]。在循环n-1次以后，如果还存在dist[v]<dist[u]+Map[u][v]，则说明有负圈。这样，我们找正圈也有方法了：dist数组初始化为负无穷。如果dist[v]>dist[u]+Map[u][v],则dist[v]=dist[u]+Map[u][v]。循环n-1次以后，如果还存在dist[v]>dist[u]+Map[u][v]，则说明有正圈。
其中，要注意的是。可以往下一房间走的条件是当前的能量值大于0。
题意：有n个房间(n<=100)，每个房间有一个点权（第1号房间和第n号房间权值均为0），到达该房间时会自动获得该点权（可能为负权）。给出一些无向边。有一个人，初始有能量值100，初始位置是第1号房间，要走到第n号房间，且路途中不得使身上能量值小于或等于0。能到达第n个房间就算赢，问能否赢。
第一行输入n，接下来n行，第一个输入权值，第二个输入表示从这个房间可到达m个房间，接下来输入m个可到达的房间。
题意：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #define M 105 5 #define inf 0x3f3f3f3f 6 using namespace std; 7 struct node 8 { 9 int a,b; 10 } ss[M*M]; 11 int map[M][M],pow[M],n; 12 13 void floyd() //判断是否可以从起点走到终点 14 { 15 int i,j,k; 16 for(k=1; k<=n; k++) 17 for(i=1; i<=n; i++) 18 for(j=1; j<=n; j++) 19 if(!map[i][j]) 20 map[i][j]=map[i][k]&&map[k][j]; 21 } 22 23 int bellman_ford(int len) //判断是否有正环，如果有，且与n是连通的，则输出可以到达 24 { 25 int dis[105]; 26 int i,j,k,a,b; 27 for(i=1; i<=n; i++) 28 dis[i]=-inf; 29 dis[1]=100; 30 for(k=1; k<n; k++) //多次循环，一种算法 31 { 32 for(i=0; i<len; i++) 33 { 34 a=ss[i].a; 35 b=ss[i].b; 36 if(dis[b]<dis[a]+pow[b] && dis[a]+pow[b]>0) 37 dis[b]=dis[a]+pow[b]; 38 } 39 } 40 41 for(i=0; i<len; i++) 42 { 43 a=ss[i].a; 44 b=ss[i].b; 45 if(dis[b]<dis[a]+pow[b] && dis[a]+pow[b]>0 && map[b][n]) 46 return 1; 47 } 48 return dis[n]>0; 49 } 50 51 int main() 52 { 53 int i,j,len,m,x; 54 while(~scanf("%d",&n)) 55 { 56 if(n==-1) break; 57 memset(ss,0,sizeof(ss)); 58 memset(pow,0,sizeof(pow)); 59 memset(map,0,sizeof(map)); 60 len=0; 61 for(i=1; i<=n; i++) 62 { 63 scanf("%d%d",&pow[i],&m); 64 for(j=0; j<m; j++) 65 { 66 scanf("%d",&x); 67 ss[len].a=i; 68 ss[len].b=x; 69 map[i][x]=1; 70 len++; 71 } 72 } 73 floyd(); 74 if(!map[1][n]) //若不能到终点，直接输出不能 75 { 76 printf("hopeless\n"); 77 continue; 78 } 79 if(bellman_ford(len)) 80 printf("winnable\n"); 81 else 82 printf("hopeless\n"); 83 } 84 return 0; 85 }
