hdoj 1317 XYZZY 【spfa判断正环求最长路径&&floyd求传递闭包】
XYZZY
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3174 Accepted Submission(s): 880
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
题目意思：有n个房间，每个房间都有一个能量值（可正可负）和对应的可以离开该房间的道路，离开这个房间需要在原有能量值的基础上加上该房间的能量值，当已有能量为0时游戏失败，问能否从房间1到达房间n。 起始有100个能量值。
条件：1，假设存在一个房间mid，使得1->mid->n 有正环且三者连通，那么一定可以到达第n个房间；
2，用spfa求最长路径，在不存在正环的情况下到达n的最长路径大于0说明可以到达第n个房间。
实现：先用floyd-warshall算法求传递闭包 map[ i ] [ j ] = map[ i ] [ j ] || (map[ i ][ k ] & map[ k ][ j ])。再用spfa求最长路径，在该过程中碰到正环中的点，对其判断（条件1），若成功直接跳出，反之 不在清除标记继续下一个（不让其进队列）。队列为空时，若条件1没能成立，那么说明没有正环 或者 存在的正环对1->n过程无法产生影响，我们只需判断1->n的最长路径是否大于0即可。
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#define MAXN 100+10
#define MAXM 10000+10
#define INF 10000000+10
using namespace std;
struct Edge
{
int to, val, next;
}edge[MAXM];
int n;
int dist[MAXN], vis[MAXN], used[MAXN], head[MAXN], top;
queue<int> Q;
int map[MAXN][MAXN];
void init()
{
top = 0;
memset(map, 0, sizeof(map));
for(int i = 1; i <= n; i++)
{
vis[i] = 0;
used[i] = 0;
head[i] = -1;
}
}
void addedge(int a, int b, int d)
{
edge[top].to = b;
edge[top].val = d;
edge[top].next = head[a];
head[a] = top++;
}
void getmap()
{
int x, y, d, num;
for(x = 1; x <= n; x++)
{
scanf("%d%d", &d, &num);
while(num--)
{
scanf("%d", &y);
addedge(x, y, d);
map[x][y] = 1;
}
}
}
int floyd()//求传递闭包
{
int k, i, j;
for(k = 1; k <= n; k++)
{
for(i = 1; i <= n; i++)
{
for(j = 1; j <= n; j++)
map[i][j] = map[i][j] || (map[i][k]&map[k][j]);
}
}
}
int judge(int mid)
{
if(map[1][mid] && map[mid][n])
return 1;
else
return 0;
}
int spfa()//求最长路径 判断是否存在正环
{
int i, j;
for(i = 1; i <= n; i++)
dist[i] = 0;
while(!Q.empty())
{
Q.pop();
}
dist[1] = 100;
Q.push(1);
vis[1] = 1;
used[1]++;
while(!Q.empty())
{
int u = Q.front();
Q.pop();
if(used[u] > n)//找到正环中的点
{
if(judge(u))//能通过该点到达终点
return 1;
else
continue;
}
vis[u] = 0;
for(i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(dist[v] < dist[u] + edge[i].val)
{
dist[v] = dist[u] + edge[i].val;
if(!vis[v])
{
vis[v] = 1;
used[v]++;
Q.push(v);
}
}
}
}
if(dist[n] > 0)//可以到达终点
return 1;
else
return 0;
}
int main()
{
while(scanf("%d", &n), n != -1)
{
init();
getmap();
floyd();
if(spfa())
printf("winnable\n");
else
printf("hopeless\n");
}
return 0;
}
看了其他人的题解，思路有一点不同： 在碰到正环就跳出，遍历所有点判断（条件1）是否成立。
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#define MAXN 100+10
#define MAXM 10000+10
#define INF 10000000+10
using namespace std;
struct Edge
{
int to, val, next;
}edge[MAXM];
int n;
int dist[MAXN], vis[MAXN], used[MAXN], head[MAXN], top;
queue<int> Q;
int map[MAXN][MAXN];
void init()
{
top = 0;
memset(map, 0, sizeof(map));
for(int i = 1; i <= n; i++)
{
vis[i] = 0;
used[i] = 0;
head[i] = -1;
}
}
void addedge(int a, int b, int d)
{
edge[top].to = b;
edge[top].val = d;
edge[top].next = head[a];
head[a] = top++;
}
void getmap()
{
int x, y, d, num;
for(x = 1; x <= n; x++)
{
scanf("%d%d", &d, &num);
while(num--)
{
scanf("%d", &y);
addedge(x, y, d);
map[x][y] = 1;
}
}
}
int floyd()//求传递闭包
{
int k, i, j;
for(k = 1; k <= n; k++)
{
for(i = 1; i <= n; i++)
{
for(j = 1; j <= n; j++)
map[i][j] = map[i][j] || (map[i][k]&map[k][j]);
}
}
}
int judge(int mid)
{
if(map[1][mid] && map[mid][n])
return 1;
else
return 0;
}
int spfa()//求最长路径 判断是否存在正环
{
int i, j;
for(i = 1; i <= n; i++)
dist[i] = 0;
while(!Q.empty())
{
Q.pop();
}
dist[1] = 100;
Q.push(1);
vis[1] = 1;
used[1]++;
while(!Q.empty())
{
int u = Q.front();
Q.pop();
vis[u] = 0;
if(used[u] > n)
break;
for(i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(dist[v] < dist[u] + edge[i].val)
{
dist[v] = dist[u] + edge[i].val;
if(!vis[v])
{
vis[v] = 1;
used[v]++;
Q.push(v);
}
}
}
}
if(dist[n] > 0)//可以到达终点
return 1;
else
{
for(i = 1; i <= n; i++)
{
if(used[i] > n && map[1][i] && map[i][n])
return 1;
}
return 0;
}
}
int main()
{
while(scanf("%d", &n), n != -1)
{
init();
getmap();
floyd();
if(spfa())
printf("winnable\n");
else
printf("hopeless\n");
}
return 0;
}