poj 1860 最短路

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2015/1/26

算法不难，主要别用矩阵去建边就好= =用矩阵很烦（没写出来）。

#include<map>#include<queue>#include<stack>#include<cmath>#include<vector>#include<climits>#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;  #define mod 10007#define lson pos<<1,l,mid#define sc(n) scanf("%d",&n)#define rson pos<<1|1,mid+1,r#define pr(n) printf("%d\n",n)#define met(n,m) memset(n, m, sizeof(n))#define F(x,y,i) for(int i = x;i > y; i--)#define f(x,y,i) for(int i = x;i < y; i++)#define ff(x,y,i) for(int i = x;i <= y; i++)#define FF(x,y,i) for(int i = x;i >= y; i--) const int N=100500;const int inf = INT_MAX;int Max(int a,int b){return a>b?a:b;}int Min(int a,int b){return  a<b?a:b;}int n, m, tot;struct node{int from,to;double rate; double commission;}s[1000];double dis[1000];void BF(int x,double num){met(dis,0);dis[x]= num;f(1,n,i){int temp = 1;f(0,tot,j){if(dis[s[j].to] < (dis[s[j].from] - s[j].commission)* s[j].rate){dis[s[j].to] = (dis[s[j].from] - s[j].commission)* s[j].rate;temp =  0;}}if(temp)break;}f(0,tot,i){if(dis[s[i].to] < (dis[s[i].from] - s[i].commission)* s[i].rate){printf("YES\n");return ;}}printf("NO\n");return ;}int main()  {      int a, b , x;    double num, r, c;    while(~scanf("%d%d%d%lf",&n,&m,&x,&num))    {    tot = 0;    f(0,m,i)    {    scanf("%d%d%lf%lf",&a,&b,&r,&c);    s[tot].from = a;    s[tot].to = b;    s[tot].rate = r;    s[tot++].commission = c;    scanf("%lf%lf",&r,&c);    s[tot].from = b;    s[tot].to = a;    s[tot].rate = r;    s[tot++].commission = c;     }        BF(x, num);        }    return 0;  }

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