Hdu 1531(差分约束)

Hdu 1531
题目很烦，慢慢体会。
1、SPFA的写法
2、SPFA判负环搞清楚有几个点
3、最短(长)路中的dis[i]表示 a[i]-a[0]的最大(小)值

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define ms(i,j) memset(i, j, sizeof i);

const int MAXN = 100 + 5;

struct edge
{
    int u, v, c;
}e[MAXN*MAXN]; 
int en;

struct sysdc
{
    int dis[MAXN];
    int vi[MAXN];
    int cir[MAXN];
    vector<int> G[MAXN];
    int n;
    void init(int n)
    {
        this->n = n;
        for (int i=0;i<=n;i++)
        {
            G[i].clear();
        }
    }
    int addedge(int a, int b, int c)//添加a-b>=c 
    {
        en++;
        e[en].u = b;
        e[en].v = a;
        e[en].c = c;
        G[b].push_back(en);
    }
    int solve()
    {
        queue<int> q;
        for (int i=0;i<=n;i++)
        {
            vi[i] = false;
            cir[i] = 1;
            dis[i] = 0;
            q.push(i);
        }
        while (!q.empty())
        {
            int p = q.front(); q.pop();
            vi[p] = false;
            for (int i=0;i<G[p].size();i++)
            {
                edge ed = e[G[p][i]];
                if (dis[p]+ed.c>dis[ed.v])
                {
                    dis[ed.v] = dis[p]+ed.c;
                    if (!vi[ed.v])
                    {
                        vi[ed.v] = true;
                        cir[ed.v]++;
                        if (cir[ed.v]>n+1) return false;
                        q.push(ed.v);
                    }
                }
            }
        }
        return true;
    }
}sdc;

int n, m;

void init()
{
    en = 0;
    sdc.init(n); 
    for (int i=1;i<=m;i++)
    {
        int si, ni, ki;
        char oi, sb;
        scanf("%d %d %c%c %d", &si, &ni, &oi, &sb, &ki);
        if (oi=='g')//>
        {
            sdc.addedge(si-1, si+ni, ki+1);
        } else sdc.addedge(si+ni, si-1, -ki+1);//<
    }
}
void solve()
{
    if (sdc.solve())
    {
        printf("lamentable kingdom\n"); 
    } else printf("successful conspiracy\n"); 
}
int main()
{
    while (scanf("%d%d", &n, &m)==2&&n&&m)
    {
        init();
        solve();
    }
    return 0;
}