|hdu 1531|差分约束|King

Hdu传送门
题目很烦，慢慢体会。

/*     Hdu 1531     差分约束系统     教训：    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;}