UVa:10000 Longest Paths (DAG上的最长路)

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DAG上的最长路。

如果使用floyd可以将所有边值取反，这样就转化成了最短路问题了。

用时0.176s。

#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <algorithm>#define MAXN 105using namespace std;int main(){    int n,s,kase=0;    while(scanf("%d",&n)&&n)    {        int gl[MAXN][MAXN];        scanf("%d",&s);        int p,q;        memset(gl,0x7f,sizeof(gl));        int INF=gl[0][0];        while(scanf("%d%d",&p,&q)&&!(!p&&!q))            gl[p][q]=-1;        for(int i=0; i<=n; ++i)            gl[i][i]=0;        for(int k=1; k<=n; ++k)            for(int i=1; i<=n; ++i)                for(int j=1; j<=n; ++j)                    if(gl[i][k]<INF&&gl[k][j]<INF)                        gl[i][j]=min(gl[i][j],gl[i][k]+gl[k][j]);        int ans=0,ed=0;        for(int i=1; i<=n; ++i)            if(gl[s][i]<ans)            {                ans=gl[s][i];                ed=i;            }        printf("Case %d: The longest path from %d has length %d, finishing at %d.\n",++kase,s,-ans,ed);        printf("\n");    }    return 0;}

用邻接矩阵实现的SPFA。

用时0.056s。

#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#define MAXN 105using namespace std;int main(){    int n,s,kase=0;    while(scanf("%d",&n)&&n)    {        int gl[MAXN][MAXN]= {0};        scanf("%d",&s);        int p,q;        while(scanf("%d%d",&p,&q)&&!(!p&&!q))            gl[p][q]=-1;        for(int i=0; i<=n; ++i) gl[i][i]=0;        int dis[MAXN];        memset(dis,0x7f,sizeof(dis));        int INF=dis[0];        queue<int> que;        que.push(s);        dis[s]=0;        bool inq[MAXN]= {0};        inq[s]=true;        while(!que.empty())        {            int getfront=que.front();            que.pop();            for(int i=1; i<=n; ++i)            {                if(gl[getfront][i]&&dis[getfront]<INF&&dis[getfront]+gl[getfront][i]<dis[i])                {                    dis[i]=dis[getfront]+gl[getfront][i];                    if(!inq[i])                    {                        que.push(i);                        inq[i]=true;                    }                }            }            inq[getfront]=false;        }        int ans=0,ed=0;        for(int i=1; i<=n; ++i)            if(dis[i]<ans)            {                ans=dis[i];                ed=i;            }        printf("Case %d: The longest path from %d has length %d, finishing at %d.\n",++kase,s,-ans,ed);        printf("\n");    }    return 0;}

最后这个是邻接表实现的SPFA。

用时0.038s。

#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#define MAXN 105using namespace std;struct Edge{    int to,cost;    Edge(int a,int b):to(a),cost(b) {}};int main(){    int n,s,kase=0;    while(scanf("%d",&n)&&n)    {        vector<Edge> gl[MAXN];        scanf("%d",&s);        int p,q;        while(scanf("%d%d",&p,&q)&&!(!p&&!q))            gl[p].push_back(Edge(q,1));        int dis[MAXN]= {0};        queue<int> que;        que.push(s);        bool inq[MAXN]= {0};        inq[s]=true;        while(!que.empty())        {            int getfront=que.front();            que.pop();            for(int i=0; i<gl[getfront].size(); ++i)            {                int t=gl[getfront][i].to,w=gl[getfront][i].cost;                if(dis[getfront]+w>dis[t])                {                    dis[t]=dis[getfront]+w;                    if(!inq[t])                    {                        que.push(t);                        inq[t]=true;                    }                }            }            inq[getfront]=false;        }        int ans=0,ed=0;        for(int i=1; i<=n; ++i)            if(dis[i]>ans)            {                ans=dis[i];                ed=i;            }        printf("Case %d: The longest path from %d has length %d, finishing at %d.\n",++kase,s,ans,ed);        printf("\n");    }    return 0;}

用Bellmanford可以直接改变松弛条件实现求最长路，floyd似乎不行，得将权值取反转化成最短路求解。