poj 1860&poj2240 似负环

来源：互联网发布：windows如何设置锁屏编辑：程序博客网时间：2024/06/16 20:01

poj1860:从自己的币种出发，如果发现正环，那么则可以不停地走环路以增加自己手中的价值.题目转换成寻找正环。

poj2240:遍历每一个点，设币为1元，如果发现正环，则可以增长前。两种代码spfa&floyd

poj1860:

#include <map>#include <set>#include <queue>#include <stack>#include <math.h>#include <vector>#include <cstdio>#include <string>#include<string.h>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;int n,m,s;double v;double c[110][110],r[110][110];inline void read(int &m){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}m=x*f;}bool spfa(){double d[110];for(int i=1;i<=n;i++) d[i]=-111111;d[s]=v;queue<int>que;que.push(s);int cnt=0;while(!que.empty()){if(cnt>n*n)return 1;int v=que.front();que.pop();cnt++;for(int i=1;i<=n;i++)if(r[v][i]>=0&&(d[v]-c[v][i])*r[v][i]>d[i]){d[i]=(d[v]-c[v][i])*r[v][i];que.push(i);}}return  0;}int main(){cin>>n>>m>>s>>v;for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)r[i][j]=-1.0;for(int i=1;i<=m;i++){int a,b;double c1,r1,c2,r2;cin>>a>>b>>r1>>c1>>r2>>c2;c[a][b]=c1;c[b][a]=c2;r[a][b]=r1;r[b][a]=r2;}if(spfa())cout<<"YES\n";elsecout<<"NO\n";return 0;}

poj2240:

spfa

#include <map>#include <set>#include <queue>#include <stack>#include <math.h>#include <vector>#include <cstdio>#include <string>#include<string.h>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;map<string,int>mm;int n,m;double graph[40][40];bool spfa(int s){double d[40];for(int i=1;i<=n;i++) d[i]=-111111;d[s]=1;int cnt=0;queue<int>que;que.push(s);while(!que.empty()){if(cnt++>n*n)return 1;int v=que.front();que.pop();for(int i=1;i<=n;i++)if(graph[v][i]>=0&&d[i]<d[v]*graph[v][i]){d[i]=d[v]*graph[v][i];que.push(i);}}return 0;}int main(){int num=1;while(scanf("%d",&n),n){memset(graph,-1.0,sizeof(graph));for(int i=1;i<=n;i++){string st;cin>>st;mm[st]=i;}scanf("%d",&m);for(int i=1;i<=m;i++){string st1,st2;double d;cin>>st1>>d>>st2;graph[mm[st1]][mm[st2]]=d;}printf("Case %d: ",num++);int i;for(i=1;i<=n;i++)if(spfa(i)){printf("Yes\n");break;}if(i>n)printf("No\n");}return 0;}

floyd:

#include <map>#include <set>#include <queue>#include <stack>#include <math.h>#include <vector>#include <cstdio>#include <string>#include<string.h>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;map<string,int>mm;int n,m;double graph[40][40];void floyd(){for(int k=1;k<=n;k++)for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(graph[i][k]*graph[k][j]>graph[i][j])graph[i][j]=graph[i][k]*graph[k][j];}int main(){int num=1;while(scanf("%d",&n),n){memset(graph,0,sizeof(graph));for(int i=1;i<=n;i++){string st;cin>>st;mm[st]=i;graph[i][i]=1;}scanf("%d",&m);for(int i=1;i<=m;i++){string st1,st2;double d;cin>>st1>>d>>st2;graph[mm[st1]][mm[st2]]=d;}floyd();bool ok=0;printf("Case %d: ",num++);for(int i=1;i<=n;i++)if(graph[i][i]>1){printf("Yes\n");ok=1;break;}if(ok) continue;printf("No\n");}return 0;}

0 0