首页 > 代码库 > CodeForces 721C Journey

CodeForces 721C Journey

$dp$,拓扑排序。

记$dp[i][j]$表示走到节点$i$,走过了$j$个点的最小时间,然后就可以递推了。要注意的是节点$1$的入度一开始不一定等于$0$。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<iostream>using namespace std;typedef long long LL;const double pi=acos(-1.0),eps=1e-6;void File(){    freopen("D:\\in.txt","r",stdin);    freopen("D:\\out.txt","w",stdout);}template <class T>inline void read(T &x){    char c = getchar();    x = 0;    while(!isdigit(c)) c = getchar();    while(isdigit(c)) { x = x * 10 + c - 0; c = getchar(); }}const int INF=0x7fffffff;const int maxn=5010;int dp[maxn][maxn];struct Edge{    int v,nx; int w;}e[maxn];int h[maxn],sz,r[maxn];int n,m,T;queue<int>Q;int pre[maxn][maxn];void add(int u,int v,LL w){    e[sz].v=v; e[sz].w=w;    e[sz].nx=h[u]; h[u]=sz++;}int main(){    scanf("%d%d%d",&n,&m,&T);    for(int i=0;i<=n;i++)    {        h[i]=-1;        for(int j=0;j<=n;j++)        {            dp[i][j]=INF;            pre[i][j]=-1;        }    }    for(int i=1;i<=m;i++)    {        int u,v; LL w; scanf("%d%d%d",&u,&v,&w);        add(u,v,w); r[v]++;    }    dp[1][1]=0;    for(int i=1;i<=n;i++) if(r[i]==0) Q.push(i);    bool flag=0;    while(!Q.empty())    {        int top=Q.front(); Q.pop();        if(top==1) flag=1;        if(flag==0)        {            for(int i=h[top];i!=-1;i=e[i].nx)            {                int to=e[i].v;                r[to]--;                if(r[to]==0) Q.push(to);            }            continue;        }        for(int i=h[top];i!=-1;i=e[i].nx)        {            int to=e[i].v;            for(int j=1;j<=n;j++)            {                if(dp[top][j-1]==INF) continue;                if(dp[top][j-1]+e[i].w>T) continue;                if(dp[top][j-1]+e[i].w>=dp[to][j]) continue;                pre[to][j]=(top-1)*n+j-1-1;                dp[to][j]=dp[top][j-1]+e[i].w;            }            r[to]--;            if(r[to]==0) Q.push(to);        }    }    int sum;    for(int i=1;i<=n;i++) if(dp[n][i]<=T) sum=i;    cout<<sum<<endl;    int nowx=n,nowy=sum; stack<int>S;    while(1)    {        S.push(nowx);        int tx,ty;        tx=pre[nowx][nowy]/n; tx++;        ty=pre[nowx][nowy]%n; ty++;        if(pre[nowx][nowy]==-1) break;        nowx=tx; nowy=ty;    }    while(!S.empty())    {        cout<<S.top()<<" "; S.pop();    }    return 0;}

 

CodeForces 721C Journey