Codeforces Round #374 (Div. 2)

记录一个菜逼的成长。。

A,B水题。
C. Journey
题目大意：
给你N个点M条边的DAG，有T的时间限制.接下来M行(u,v,t）代表从u到v的时间为t。你站在1点 ，问到n点，不超过T时间最多走过几个点，并输出路径。

跟拓扑排序很像，加上dp。
dp[i][j] := 站在i点，走过j个点的最短时间。
显然dp[i][j] = min(dp[i][j] + t[i],dp[to[i]][j+1]);
path[i][j] := 站在i点，走过j个点的前面那个点。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>#include <vector>#include <set>#include <map>#include <queue>#include <stack>#include <list>#include <deque>#include <cctype>#include <bitset>#include <cmath>using namespace std;#define ALL(v) (v).begin(),(v).end()#define cl(a,b) memset(a,b,sizeof(a))#define bp __builtin_popcount#define pb push_back#define fin freopen("D://in.txt","r",stdin)#define fout freopen("D://out.txt","w",stdout)#define lson t<<1,l,mid#define rson t<<1|1,mid+1,r#define seglen (node[t].r-node[t].l+1)#define pi 3.1415926#define e  2.718281828459typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> PII;typedef pair<LL,LL> PLL;typedef vector<PII> VPII;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;template <typename T>inline void read(T &x){    T ans=0;    char last=' ',ch=getchar();    while(ch<'0' || ch>'9')last=ch,ch=getchar();    while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();    if(last=='-')ans=-ans;    x = ans;}inline bool DBread(double &num){    char in;double Dec=0.1;    bool IsN=false,IsD=false;    in=getchar();    if(in==EOF) return false;    while(in!='-'&&in!='.'&&(in<'0'||in>'9'))        in=getchar();    if(in=='-'){IsN=true;num=0;}    else if(in=='.'){IsD=true;num=0;}    else num=in-'0';    if(!IsD){        while(in=getchar(),in>='0'&&in<='9'){            num*=10;num+=in-'0';}    }    if(in!='.'){        if(IsN) num=-num;            return true;    }else{        while(in=getchar(),in>='0'&&in<='9'){                num+=Dec*(in-'0');Dec*=0.1;        }    }    if(IsN) num=-num;    return true;}template <typename T>inline void write(T a) {    if(a < 0) { putchar('-'); a = -a; }    if(a >= 10) write(a / 10);    putchar(a % 10 + '0');}/******************head***********************/const int maxn = 5000 + 10;int head[maxn],cnt;int path[maxn][maxn],dp[maxn][maxn],indegree[maxn];int n,m,T;struct Edge{    int to,next,w;}edge[maxn];void add(int u,int v,int w){    edge[cnt].to = v;    edge[cnt].w = w;    edge[cnt].next = head[u];    head[u] = cnt++;}void init(){    cl(head,-1);    cl(path,0);    cl(indegree,0);    for( int i = 1; i <= n; i++ ){        for( int j = 1; j <= n; j++ ){            dp[i][j] = INF;        }    }}void print_path(int num){    stack<int>s;    int i = n,j = num;    while(path[i][j]){        s.push(i);        i = path[i][j];        j--;    }    int flag = 1;    s.push(1);    while(!s.empty()){        if(flag){            printf("%d",s.top());            s.pop();            flag = 0;        }        else {            printf(" %d",s.top());            s.pop();        }    }    puts("");}void solve(){    queue<int>q;    dp[1][1] = 0;    for( int i = 1; i <= n; i++ )        if(!indegree[i])q.push(i);    while(!q.empty()){        int f = q.front();q.pop();        for( int i = head[f]; ~i; i = edge[i].next){            int v = edge[i].to,w = edge[i].w;            for( int j = 1; j <= n; j++ ){                if(dp[f][j] + w < dp[v][j+1]){                    dp[v][j+1] = dp[f][j] + w;                    path[v][j+1] = f;                }            }            if(!--indegree[v])                q.push(v);        }    }    for( int i = n; i > 0; i-- ){        if(dp[n][i] <= T){            printf("%d\n",i);            print_path(i);            break;        }    }}int main(){    while(~scanf("%d%d%d",&n,&m,&T)){        init();        for( int i = 0; i < m; i++ ){            int u,v,t;            scanf("%d%d%d",&u,&v,&t);            add(u,v,t);            indegree[v]++;        }        solve();    }    return 0;}