POJ 1364 King 差分约束 找负环

嘛，虽然是一道水题+模板题，不过还是学到了很多东西的，记录一下。

首先题目给出的不等式是小于，但是差分约束系统只能处理小于等于的情况，所以要转化成小于等于的进行处理。对于整数处理方法非常简单= =

然后是找负环的情况，其实不需要考虑图连不连通，只要一开始就把所有的点的d置成0，然后都push进队列里面就好了。

PS：这种方法同样可以用在处理多源点最短路问题上。

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#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack>   using namespace std;   typedef long long LL; const int maxn = 105; const int maxm = 5005; int first[maxn],nxt[maxm],v[maxm],w[maxm]; int d[maxn],n,m,ecnt,cnt[maxn]; bool inq[maxn];   void adde(int _u,int _v,int _w) {     v[ecnt] = _v; w[ecnt] = _w;     nxt[ecnt] = first[_u];     first[_u] = ecnt;     ecnt++; }   void solve() {     bool bad = false;     queue<int> q;     for(int i = 0;i <= n;i++) {         d[i] = 0;         cnt[i] = 1;         inq[i] = true;         q.push(i);     }     while(!q.empty() && !bad) {         int x = q.front(); q.pop();         inq[x] = false;         for(int i = first[x];i != -1;i = nxt[i]) {             if(d[v[i]] > d[x] + w[i]) {                 if(!inq[v[i]]) {                     inq[v[i]] = true;                     cnt[v[i]]++;                     q.push(v[i]);                     if(cnt[v[i]] > n + 2) {                         bad = true;                         break;                     }                 }                 d[v[i]] = d[x] + w[i];             }         }     }     if(bad) puts("successful conspiracy");     else puts("lamentable kingdom"); }   int main() {     while(scanf("%d",&n),n) {         ecnt = 0;         char sig[16];         scanf("%d",&m);         memset(first,-1,sizeof(first));         memset(nxt,-1,sizeof(nxt));         for(int i = 0;i < m;i++) {             int a,b,c;             scanf("%d %d %s %d",&a,&b,sig,&c);             if(sig[0] == 'g') adde(b + a,a - 1,-c - 1);             else adde(a - 1,b + a,c - 1);         }         /*         for(int i = 1;i <= n;i++) {             adde(i,i - 1,0);         }         */         solve();     }     return 0; }