poj3621 Sightseeing Cows bellman-ford 判负权环

Sightseeing Cows

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6908 Accepted: 2291


Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free time.

Fortunately, they have a detailed city map showing the L (2 ≤ L ≤ 1000) major landmarks (conveniently numbered 1.. L) and the P (2 ≤ P ≤ 5000) unidirectional cow paths that join them. Farmer John will drive the cows to a starting landmark of their choice, from which they will walk along the cow paths to a series of other landmarks, ending back at their starting landmark where Farmer John will pick them up and take them back to the farm. Because space in the city is at a premium, the cow paths are very narrow and so travel along each cow path is only allowed in one fixed direction.

While the cows may spend as much time as they like in the city, they do tend to get bored easily. Visiting each new landmark is fun, but walking between them takes time. The cows know the exact fun values Fi (1 ≤ Fi ≤ 1000) for each landmark i.

The cows also know about the cowpaths. Cowpath i connects landmark L1i to L2i (in the direction L1i -> L2i ) and requires time Ti (1 ≤ Ti ≤ 1000) to traverse.

In order to have the best possible day off, the cows want to maximize the average fun value per unit time of their trip. Of course, the landmarks are only fun the first time they are visited; the cows may pass through the landmark more than once, but they do not perceive its fun value again. Furthermore, Farmer John is making the cows visit at least two landmarks, so that they get some exercise during their day off.

Help the cows find the maximum fun value per unit time that they can achieve.


* Line 1: Two space-separated integers: L and P
* Lines 2..L+1: Line i+1 contains a single one integer: Fi
* Lines L+2..L+P+1: Line L+i+1 describes cow path i with three space-separated integers: L1i , L2i , and Ti


* Line 1: A single number given to two decimal places (do not perform explicit rounding), the maximum possible average fun per unit time, or 0 if the cows cannot plan any trip at all in accordance with the above rules.

Sample Input

5 7
1 2 3
2 3 2
3 4 5
3 5 2
4 5 5
5 1 3
5 2 2

Sample Output



牛可以从任意点出发, 每个点有欢乐值, 一个点可以去多次, 但是欢乐值只增加一次, 每条边有时间消耗, 求一条回路使得 总欢乐值/总时间 最大


/** head-file **/

#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>

/** define-for **/

#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)

/** define-useful **/

#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair

/** test **/

#define Display(A, n, m) {                      \
    REP(i, n){                                  \
        REP(j, m) cout << A[i][j] << " ";       \
        cout << endl;                           \
    }                                           \

#define Display_1(A, n, m) {                    \
    REP_1(i, n){                                \
        REP_1(j, m) cout << A[i][j] << " ";     \
        cout << endl;                           \
    }                                           \

using namespace std;

/** typedef **/

typedef long long LL;

/** Add - On **/

const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };

const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-6;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;

const int maxn=111111;
const int maxm=111111;
int A[maxn];
int B[maxn];
int cost[maxn];
double E[maxm];
double EE[maxm];
double dis[maxn];
int n,m;

bool init()
    if (scanf("%d%d",&n,&m)==EOF) return false;
    REP_1(i,n) scanf("%d",&cost[i]);
    return true;

bool bellman_ford()
    bool flag;
        REP_1(j,m) if (dis[A[j]]+E[j]<dis[B[j]])
        if (!flag) return true;
    return false;

void solve()
    double l,r,mid;
    while (l+EPS<r)
        REP_1(i,m) E[i]=EE[i]*mid-cost[B[i]];
        if (bellman_ford()) r=mid;
        else l=mid;
int main()
    while (init())
    return 0;
    原文作者:Bellman - ford算法
    原文地址: https://blog.csdn.net/cyendra/article/details/9816815