UVA11478 Halum (差分约束)

讲每个点的所有操作同一起来， 设sum[u]为u节点上d值之和。二分枚举答案x，就得到了  sum(a)-sum(b)+w(a,b)>=x ，即sum[b] - sum[a] <= w[a, b] - x.对于差分约束系统来讲，j - i <= w， 建边i->j，权值为w。

建好图后，题意要求最小边最大且大于0.那么如果最小边为原图最大边+1的时候任不存在负环，那么答案显然是inf了。而如果原图存在权值和<=0的环，那么答案就是no solution了。

而且这个题有点猥琐。。。spfa中，判负环的节点入队限制如果为n的话，用vector的spfa模板是会TLE的，但是改成sqrt(n)后，100+ms无压力。。。

#include<algorithm>#include<iostream>#include<cstring>#include<cstdlib>#include<fstream>#include<sstream>#include<string>#include<vector>#include<bitset>#include<cstdio>#include<queue>#include<stack>#include<cmath>#include<map>#include<set>using namespace std;#define FF(i, a, b) for(int i=a; i<b; i++)#define FD(i, a, b) for(int i=a; i>=b; i--)#define REP(i, n) for(int i=0; i<n; i++)#define CLR(a, b) memset(a, b, sizeof(a))#define debug puts("**debug**")#define LL long long#define PB push_backconst int maxn = 555;const int INF = 100000;int n, m, d[maxn], cnt[maxn], limit;bool inq[maxn];struct Edge{    int from, to, dist;};vector<Edge> edges;vector<int> G[maxn];void add(int a, int b, int c){    edges.PB((Edge){a, b, c});    int nc = edges.size();    G[a].PB(nc-1);}inline void init(){    REP(i, n+1) G[i].clear(); edges.clear();    FF(i, 1, n+1) add(0, i, 0);    limit = (int) sqrt (n + 0.0) + 1;}bool negacycle(){    queue<int> q; q.push(0); d[0] = 0;    CLR(inq, 0); CLR(cnt, 0);    FF(i, 1, n+1) d[i] = INF;    while(!q.empty())    {        int u = q.front(); q.pop();        inq[u] = false;        int nc = G[u].size();        REP(i, nc)        {            Edge e = edges[G[u][i]];            if(d[e.to] > d[u] + e.dist)            {                d[e.to] = d[u] + e.dist;                if(!inq[e.to])                {                    q.push(e.to);                    inq[e.to] = 1;                    if(++cnt[e.to] > limit) return false;                }            }        }    }    return true;}bool ok(int x){    int nc = edges.size();    REP(i, nc) edges[i].dist -= x;    bool ret = negacycle();    REP(i, nc) edges[i].dist += x;    return ret;}int main(){    while(~scanf("%d%d", &n, &m))    {        init();        int maxv = -INF, a, b, c;        REP(i, m)        {            scanf("%d%d%d", &a, &b, &c);            add(a, b, c);            maxv = max(maxv, c);        }        if(ok(maxv + 1))    puts("Infinite");        else if(!ok(1))      puts("No Solution");        else        {            int L = 1, R = maxv, ans , M;            while(L <= R)            {                M = (L + R) >> 1;                if(ok(M)) ans = M, L = M + 1;                else R = M - 1;            }            printf("%d\n", ans);        }    }    return 0;}