hdu XYZZY-(floyd判联通+spfa判正环+点权||纯spfa边权)

题目：

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.
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.
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

题意：

一个人在1号房间有100点能量，现在他要去第n个房间，途中每个房间都有一个权值有正有负，每经过一个房间能量都要加上这个权值，如果在路上这个人的能量小于等于0了就不能在走动了。问是否能到达n号房间并且剩余能量为正。

思路：

如果途中存在正环并且这个正环中的任意一个点可以到达n房间那么这个人肯定最终能win.所以这个题就两步：
1.判断是否存在正环
2.判断每个点是否能到n点

代码1(spfa+floyd)：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int maxn = 123;
int T,n,m,flag;
int reach[maxn][maxn];
int maps[maxn][maxn];
int cnt[maxn];
int val[maxn];
int eng[maxn];

void floyd()
{
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(!reach[i][j]&&reach[i][k]&&reach[k][j])
                    reach[i][j]=1;
            }
        }
    }
}

int spfa(int st)
{
    queue<int>q;
    q.push(st);
    eng[1]=100;
    cnt[st]++;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        if(cnt[u]>n)
            return reach[u][n];
        for(int i=1;i<=n;i++)
        {
            if(maps[u][i]&&eng[i]<eng[u]+val[i])
            {
                q.push(i);
                cnt[i]++;
                eng[i]=eng[u]+val[i];
            }
        }
    }
    if(eng[n]>0)
        return true;
    else
        return false;
}

void init()
{
    memset(maps,0,sizeof(maps));
    memset(val,0,sizeof(val));
    memset(reach,0,sizeof(reach));
    memset(cnt,0,sizeof(cnt));
    memset(eng,0,sizeof(eng));
}

int main()
{
    while(scanf("%d",&n)!=EOF&&n!=-1)
    {
        init();
        for(int i=1;i<=n;i++)
        {
            int w,x;
            scanf("%d%d",&w,&m);
            val[i]=w;
            while(m--)
            {
                scanf("%d",&x);
                maps[i][x]=reach[i][x]=1;
            }
        }
        floyd();
        if(!reach[1][n])
            printf("hopeless\n");
        else
        {
            if(spfa(1))
                printf("winnable\n");
            else
                printf("hopeless\n");
        }
    }
    return 0;
}

代码2(spfa):

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int maxn = 123;
int T,n,m,tot;
int vis[maxn];
int cnt[maxn];//记录进入队列次数
int eng[maxn];//记录最大路
int head[maxn*maxn];

struct Edge
{
    int to;
    int w;
    int next;
}edge[maxn*maxn];

void add_edge(int u,int v,int w)
{
    edge[tot].to=v;
    edge[tot].w=w;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void init()
{
    tot=0;
    memset(edge,0,sizeof(edge));
    memset(vis,0,sizeof(vis));
    memset(cnt,0,sizeof(cnt));
    memset(eng,-INF,sizeof(eng));
    memset(head,-1,sizeof(head));
}

bool spfa()
{
    queue<int>q;
    q.push(1);
    eng[1]=100;
    vis[1]=1;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        cnt[u]++;
        vis[u]=0;
        if(cnt[u]>n)
        {
            eng[u]=INF;
            vis[u]=1;
        }
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            int w=edge[i].w;
            if(eng[u]+w>0&&eng[u]+w>eng[v])
            {
                eng[v]=eng[u]+w;
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
    if(eng[n]>0)
        return true;
    else
        return false;
}

int main()
{
    while(scanf("%d",&n)!=EOF&&n!=-1)
    {
        init();
        for(int i=1;i<=n;i++)
        {
            int w,x;
            scanf("%d%d",&w,&m);
            while(m--)
            {
                scanf("%d",&x);
                add_edge(i,x,w);
            }
        }
        if(spfa())
            printf("winnable\n");
        else
            printf("hopeless\n");
    }
    return 0;
}