UVA 11090 - Going in Cycle!!(最短路`Bellman-Ford)

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题目:

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=74214#problem/E

题意:

求出有向图中环的最小平均权值, 不成环则输出"No cycle found".

思路:

ford判负圈法+二分.

二分边的mid值, 若存在 c1+c2+..+ck < k*mid, (c1-mid)+(c2-mid)+...+(ck-mid) < 0,

则转化成每条边cost - mid, 判断图中是否存在负圈.

AC.

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const double esp = 1e-3;const int INF = 1e9;const int MAXM = 5005;int n, m;struct edge {    int to, fm;    double cost;}eg[MAXM];double d[55];bool find_negloop(){    memset(d, 0, sizeof(d));    for(int i = 0; i < n; ++i) {        for(int j = 0; j < m; ++j) {            edge e = eg[j];            if(d[e.to] > d[e.fm] + e.cost) {                d[e.to] = d[e.fm] + e.cost;                if(i == n-1) return true;            }        }    }    return false;}bool test(double mid){    for(int i = 0; i < m; ++i) {        eg[i].cost -= mid;    }    bool ok = find_negloop();    for(int i = 0; i < m; ++i) {        eg[i].cost += mid;    }    return ok;}int main(){//freopen("in", "r", stdin);    int T, ca = 1;    scanf("%d", &T);    while(T--) {        scanf("%d %d", &n, &m);        int u, v;        double c;        double maxn = 0;                for(int i = 0; i < m; ++i) {            scanf("%d %d %lf", &u, &v, &c);            eg[i].to = u; eg[i].fm = v; eg[i].cost = c;            maxn = max(maxn, c);        }        printf("Case #%d: ", ca++);        double l = 0, r = maxn+1;        if(!test(r)) {            printf("No cycle found.\n");            continue;        }        while(r - l > esp) {            double mid = l + (r-l) / 2.0;            //printf("%lf %lf %lf\n", l, mid, r);            if(test(mid)) r = mid;            else l = mid;        }        printf("%.2lf\n", l);    }    return 0;}

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