【差分约束】hdu 3666

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Xij * ai - L * bj >= 0

Xij * ai - U* bj <= 0

两边取对数来去除ai，bj前面的系数

有：

logbj - logai >= logXij - logU

logai - log bj <=logL - logXij

化成标准差分约束，建图，spfa，注意一个窍门，当入队总数大于2*（n+m）时就可以输出NO（因为一个点用来作为迭代点对其它点进行改进的平均次数为k，有办法证明对于通常的情况，k在2左右）

用vector实现，时间有点慢1078MS。。。

#include <list>#include <map>#include <set>#include <queue>#include <string>#include <deque>#include <stack>#include <algorithm>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <limits.h>#include <time.h>#include <string.h>using namespace std;#define LL long long#define PI acos(-1.0)#define MAX INT_MAX#define MIN INT_MIN#define eps 1e-8#define FRE freopen("a.txt","r",stdin)#define MOD 1000000007#define N 805struct node{    int to;    double w;};int n,m;double l,u;vector<node> v[N];double dis[N];bool vis[N];void build(int s,int t,double x){    node tmp;    tmp.to=t;    tmp.w=x;    v[s].push_back(tmp);}void init(){    int i,j,k;    for(i=0;i<N;i++)v[i].clear();    for(i=1;i<=n;i++){        for(j=1;j<=m;j++){            double x;            scanf("%lf",&x);            build(j+n,i,log(u/x));            build(i,j+n,log(x/l));        }    }}void spfa(){    queue<int> q;    while(!q.empty())q.pop();    int i,j,k;    memset(vis,0,sizeof(vis));    for(i=0;i<N;i++)dis[i]=MAX;    q.push(1);    dis[1]=0;    vis[1]=1;    int sum=0;    while(!q.empty()){        int top=q.front();        q.pop();        vis[top]=0;        for(i=0;i<v[top].size();i++){            int tmp=v[top][i].to;            double val=v[top][i].w;            if(dis[tmp]>dis[top]+val){                dis[tmp]=dis[top]+val;                if(!vis[tmp]){                    vis[tmp]=1;                    sum++;                    if(sum>=2*(n+m)){                        printf("NO\n");                        return ;                    }                    q.push(tmp);                }            }        }    }    printf("YES\n");    return ;}int main(){    while(scanf("%d%d%lf%lf",&n,&m,&l,&u)!=EOF){         init();         spfa();    }    return 0;}