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#PORSITELOPERDISTE | 🏛️ GOBIERNO | Puebla Nuevo Edificio de Finanzas en Puebla operará en febrero; negocian estacionamiento con BUAP (@BUAPoficial @LiliaCedillo_ ) http://dlvr.it/TQDv0v 🔗
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🏛️ GOBIERNO | Puebla Nuevo Edificio de Finanzas en Puebla operará en febrero; negocian estacionamiento con BUAP (@BUAPoficial @LiliaCedillo_ ) http://dlvr.it/TQCnhp 🔗
The ministry of our spiritual son in Bungoma has taken off like a rocket. Suddenly he's on TV, for the first time ever, touching lives. Grace of God is real. Glory to Jesus Christ forever. 👇🏽 During one on one meeting in Bungoma, my father PFA released special grace upon my life and Ministry generally. The grace is working in a powerful way. On Sunday 4/4/2021 I was to be in PEFA Satellite Nairobi to Minister, but due to lockdown, things changed. Immediately after that, I received an invitation to be the guest minister in Migori~Kehancha in a youths seminar that will begin on 10th and end on 11th of April. Today being a good Friday, I have been invited for an interview at NYOTA TV to represent the body of Christ in celebration of Easter. I call it grace at work. From Your son in Bungoma Pst Reinhard Namdi #SPFA https://www.facebook.com/161582470920698/posts/1101657830246486/ https://www.instagram.com/p/CNNPKi_JjC6/?igshid=1wum715d513e3
Mais uma vitória dos @Vikings_FA 😍❤👏👏 . Parabéns, mais uma vez, meninos! Vcs arrasam!! . #somosvikings #SOMOSMADNESS #futebolamericano #FA #FABR #SPFA #nflbrasil #nflluluzinhaclub (at Madness Fitness)
AOOO, #MADFRIENDS!! Pra quem não sabe, nós apoiamos os @Vikings_FA um time de #FutebolAmericano de #Maua! . E hoje, teve jogo amistoso, contra o @sbavengers e nós conquistamos a segunda vitória do ano. . Dizem que amistoso não vale nada, mas vale sim!! Vale para sabermos em que estamos pecando, em que estamos acertando, o que precisa ser feito e por aí vai....E aí, EU (Juzz), digo: ESSE TIME TÁ DEMAIS!! . Parabéns, meninos, puta orgulho de vcs!!! Vamos seguindo com humildade e respeito, sempre! Amo vcs! ❤✔ . . #VEMpraMADNESS #osanguefazagramacrescer #SOMOSVIKINGS #123Vikings #FA #SPFA #Fabr (at Madness Fitness)
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Fala, galera...passando pra lembrar que a seletiva dos VIKINGS é no próximo sábado!! Siga a page: @vikings_fa e inscreva-se!! . Em breve, #FlagFootball feminino! Fiquem ligadas!! . Não precisa de equipamento e nem SABER AS REGRAS...rs Podem ir assistir, tbem!! . Esperamos vcs!!! Mais informações por DM! #SOMOSVIKINGS @madnessfootballbr . . #maua #ribeiraopires #ribeirãopires #riograndedaserra #santoandre #futebolamericano #FA #SPFA #Fabr #nflbrasil (at Madness Fitness)
[BZOJ1003][ZJOI2006]物流运输trans
1003: [ZJOI2006]物流运输trans
Time Limit: 10 Sec Memory Limit: 162 MB Submit: 3869 Solved: 1589 [Submit][Status]
Description
物流公司要把一批货物从码头A运到码头B。由于货物量比较大，需要n天才能运完。货物运输过程中一般要转停好几个码头。物流公司通常会设计一条固定的运输路线，以便对整个运输过程实施严格的管理和跟踪。由于各种因素的存在，有的时候某个码头会无法装卸货物。这时候就必须修改运输路线，让货物能够按时到达目的地。但是修改路线是一件十分麻烦的事情，会带来额外的成本。因此物流公司希望能够订一个n天的运输计划，使得总成本尽可能地小。
Input
第一行是四个整数n（1<=n<=100）、m（1<=m<=20）、K和e。n表示货物运输所需天数，m表示码头总数，K表示每次修改运输路线所需成本。接下来e行每行是一条航线描述，包括了三个整数，依次表示航线连接的两个码头编号以及航线长度（>0）。其中码头A编号为1，码头B编号为m。单位长度的运输费用为1。航线是双向的。再接下来一行是一个整数d，后面的d行每行是三个整数P（ 1 < P < m）、a、b（1 < = a < = b < = n）。表示编号为P的码头从第a天到第b天无法装卸货物（含头尾）。同一个码头有可能在多个时间段内不可用。但任何时间都存在至少一条从码头A到码头B的运输路线。
Output
包括了一个整数表示最小的总成本。总成本=n天运输路线长度之和+K*改变运输路线的次数。
Sample Input
5 5 10 8 1 2 1 1 3 3 1 4 2 2 3 2 2 4 4 3 4 1 3 5 2 4 5 2 4 2 2 3 3 1 1 3 3 3 4 4 5
Sample Output
32
HINT
前三天走1-4-5，后两天走1-3-5，这样总成本为(2+2)*3+(3+2)*2+10=32
分析：SPFA+DP
先求出minPath[a,b] (1<=a<=b<=n)，代表第a天到第b天不改变路线的最短路，若minPath[a,b]==INF，代表第a天到第b天必须改变路径，也就是在这些天内关闭的码头封死了路径。
那么设DP[i]为到第i天的最小花费
DP[i] = min{DP[j] + minPath[j+1][i]*(i-j) + K} (1<=i<=n && minPath[j+1][i] != INF)
其中K代表第j天和第j+1天改变路径时的花费
DP[0] = -K (考虑更新DP[1]时候的情况)
//trans #include <stdio.h> #include <stdlib.h> #include <string.h> #include <vector> #define MAXM 1000 #define MAXN 20 #define INF 2139062143 #define MIN(a,b) ((a)<(b)?(a):(b)) using namespace std; int n,m,K,e,d; int first[MAXN+5]={0}; int next[MAXM]={0},u[MAXM]={0},v[MAXM]={0},w[MAXM]={0}; int minPath[102][102]={0}; bool close[102][25]={0}; void input() { int a,b,c; memset(first,-1,sizeof(first)); scanf("%d%d%d%d",&n,&m,&K,&e); for(int i=1;i<=e;i++) { scanf("%d%d%d",&a,&b,&c); u[i]=a;v[i]=b;w[i]=c; u[i+e]=b;v[i+e]=a;w[i+e]=c; next[i] = first[a]; first[a] = i; next[i+e] = first[b]; first[b] = i+e; } scanf("%d",&d); for(int i=1;i<=d;i++) { scanf("%d%d%d",&a,&b,&c); for(int j=b;j<=c;j++) close[j][a] = true; } } void init()//构建[i,j]天走同一条路线的最短路 { bool vis[25]={0};//为1则不能访问 bool inQ[25]={0}; int queue[200]={0}; int dis[25]={0}; int head,rear,cur; for(int a=1;a<=n;a++) { for(int b=1;b<=a;b++) { memset(vis,0,sizeof(vis)); memset(inQ,0,sizeof(inQ)); memset(dis,0x7f,sizeof(dis)); for(int k=b;k<=a;k++) for(int p=1;p<=m;p++) if(close[k][p]) vis[p] = true; head=0;rear=-1; queue[++rear] = 1; inQ[1] = true; dis[1] = 0; while(rear >= head) { cur = queue[head]; for(int k=first[cur];k!=-1;k=next[k]) { if(!vis[v[k]] && dis[v[k]] > dis[cur] + w[k]) { dis[v[k]] = dis[cur] + w[k]; if(!inQ[v[k]]) { inQ[v[k]] = true; queue[++rear] = v[k]; } } } inQ[cur] = false; head++; } minPath[b][a] = dis[m]; } } } void solve() { int dp[105]={0};//第k天的最小花费 memset(dp,0x7f,sizeof(dp)); dp[0] = -K; for(int i=1;i<=n;i++) for(int j=0;j<i;j++) if(minPath[j+1][i] != INF) dp[i] = MIN(dp[i],dp[j]+minPath[j+1][i]*(i-j)+K); printf("%d\n",dp[n]); } int main() { freopen("trans.in","r",stdin); freopen("trans.out","w",stdout); input(); init(); solve(); fclose(stdin); fclose(stdout); return 0; }