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C.Journey
读错题目了。。。不是无向图,结果建错图了(喵第4样例是变成无向就会有环的那种图)
并且这题因为要求路径点尽可能多
其实可以规约为限定路径长的拓扑排序,不一定要用最短路做
#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; }
D.Maxim and Array
没时间做。。。被第2、3题耽误了
试算了一下发现并不复杂。。。每次取绝对值最小的数(使其余值的乘积绝对值最大)这样对其加减时,总乘积变化也就最大
太多数了,直接乘会爆long long,直接判断负数个数(总乘积的正负性),以此判断要加要减
#include <stdio.h> #include <algorithm> #include <string.h> #include <iostream> #include <queue> #include <math.h> using namespace std; const int N = 200000 + 50; typedef long long ll; ll aabs(ll x) {return x<0 ? -x : x;} struct node { ll num; int id; bool operator < (const node & temp) const { return aabs(num) > aabs(temp.num); } }; ll a[N]; ll n,k,x; int main() { cin >> n >> k >> x; int sign = 1; priority_queue<node> Q; for(int i=1;i<=n;i++) { scanf("%I64d",a+i); if(a[i] < 0) sign = -sign; Q.push((node){a[i],i}); } while(k--) { node temp = Q.top();Q.pop(); if(a[temp.id] < 0) { if(sign == -1) a[temp.id] -= x; else a[temp.id] += x; if(a[temp.id] >= 0) sign = -sign; } else { if(sign == -1) a[temp.id] += x; else a[temp.id] -= x; if(a[temp.id] < 0) sign = -sign; } Q.push((node){a[temp.id],temp.id}); } for(int i=1;i<=n;i++) { printf("%I64d%c",a[i],i==n?‘\n‘:‘ ‘); } }
E.Road to Home
Codeforces Round #374 (div.2)遗憾题合集
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原文地址:http://www.cnblogs.com/dgutfly/p/5925915.html