POJ 3621 (最优比率环 二分+SPFA)

题目链接:点击这里

题意:找出一个环, 最大化∑vi∑Ei.

假设答案是ans.那么有∑vi∑Ei=ans也就是∑vi−ans∑Ei=0.因为是一个环,所以环上每个点都被算了两次,所以可以给每条边对应一个Fi表示边上两个节点权值和的一半,所以所求就是∑Fi−ans∑Ei=0.二分枚举ans, 如果找到一个负环,说明ans偏小,否则ans偏大.

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#define Clear(x,y) memset (x,y,sizeof(x))
#define FOR(a,b,c) for (int a = b; a <= c; a++)
#define REP(a,b,c) for (int a = b; a >= c; a--)
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<long long, int>
#define pb push_back
#define mod 1000000007
using namespace std;
#define maxn 1005
#define maxm 5005

int n, m;
int val[maxn];
struct node {
int u, v, num, next;
double w;
}edge[maxm];
int head[maxn], cnt;

void add_edge (int u, int v, int w) {
edge[cnt].num = w;
edge[cnt].u = u, edge[cnt].v = v, edge[cnt].next = head[u], head[u] = cnt++;
}

bool vis[maxn];
int top, num[maxn];
double d[maxn];

bool spfa (double ans) {
queue <int> q; while (!q.empty ()) q.pop ();
for (int i = 1; i <= n; i++) {
vis[i] = 1;
d[i] = 0;
q.push (i);
num[i] = 1;
}
for (int i = 0; i < cnt; i++) {
node &e = edge[i];
e.w = (val[e.u]+val[e.v])/2.0-ans*e.num;
}
while (!q.empty ()) {
int u = q.front (); q.pop ();
vis[u] = 0;
for (int i = head[u]; i+1; i = edge[i].next) {
int v = edge[i].v;
if (d[v] < d[u]+edge[i].w) {
d[v] = d[u]+edge[i].w;
if (!vis[v]) {
vis[v] = 1;
q.push (v);
if (++num[v] > n)
return 1;
}
}
}
}
return 0;
}

int main () {
//freopen ("more.in", "r", stdin);
while (scanf ("%d%d", &n, &m) == 2) {
double l = 0, r = 0;
for (int i = 1; i <= n; i++) {
scanf ("%d", &val[i]);
r += val[i];
}
Clear (head, -1);
cnt = 0;
for (int i = 1; i <= m; i++) {
int u, v, w; scanf ("%d%d%d", &u, &v, &w);
add_edge (u, v, w);
}
while (r-l > 1e-5) {
double mid = (l+r)/2;
if (spfa (mid)) l = mid;
else r = mid;
}
printf ("%.2f\n", l);
}
return 0;
}