好简单啊。。

T1裸分层图最短路。

T2裸容斥。

T3更水的DP。

代码

T1

#include <bits/stdc++.h>

#define rin(i,a,b) for(register int i=(a);i<=(b);++i)
#define irin(i,a,b) for(register int i=(a);i>=(b);--i)
#define trav(i,a) for(register int i=head[a];i;i=e[i].nxt)
typedef long long LL; using std::cin;
using std::cout;
using std::endl; inline int read(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
return x*f;
} const int MAXN=50005;
const int MAXM=100005; int n,m,k,ecnt,head[MAXN<<1];
LL dis[MAXN<<1];
bool book[MAXN<<1]; struct Edge{
int to,nxt,w;
}e[MAXM*4+MAXN]; inline void add_edge(int bg,int ed,int val){
++ecnt;
e[ecnt].to=ed;
e[ecnt].nxt=head[bg];
e[ecnt].w=val;
head[bg]=ecnt;
} std::queue<int> q; void spfa(){
while(!q.empty()) q.pop();
memset(dis,0x3f,sizeof dis);
dis[n]=0;
q.push(n);
book[n]=true;
while(!q.empty()){
int x=q.front();
trav(i,x){
int ver=e[i].to;
if(dis[ver]>dis[x]+e[i].w){
dis[ver]=dis[x]+e[i].w;
if(!book[ver]){
q.push(ver);
book[ver]=true;
}
}
}
q.pop();
book[x]=false;
}
} int main(){
n=read(),m=read(),k=read();
rin(i,1,m){
int u=read(),v=read(),w=read();
add_edge(u,v,w);
add_edge(v,u,w);
add_edge(u+n,v+n,w);
add_edge(v+n,u+n,w);
}
rin(i,1,k){
int x=read(),y=read();
add_edge(x,x+n,-y);
}
spfa();
rin(i,1,n-1){
if(dis[n+i]<=dis[i]) printf("1\n");
else printf("0\n");
}
return 0;
}

T2

#include <bits/stdc++.h>

#define rin(i,a,b) for(register int i=(a);i<=(b);++i)
#define irin(i,a,b) for(register int i=(a);i>=(b);--i)
#define trav(i,a) for(register int i=head[a];i;i=e[i].nxt)
typedef long long LL; using std::cin;
using std::cout;
using std::endl; inline int read(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
return x*f;
} typedef std::pair<int,int> tw;
typedef std::pair<int,tw> tr;
typedef std::pair<tw,tw> fo;
typedef std::pair<tw,tr> fi; inline tw mktw(int x,int y){
return (tw){x,y};
} inline tr mktr(int x,int y,int z){
return (tr){x,mktw(y,z)};
} inline fo mkfo(int x,int y,int z,int v){
return (fo){mktw(x,y),mktw(z,v)};
} inline fi mkfi(int x,int y,int z,int v,int w){
return (fi){mktw(x,y),mktr(z,v,w)};
} int n;
int mp1[1000005];
std::map<tw,int> mp2;
std::map<tr,int> mp3;
std::map<fo,int> mp4;
std::map<fi,int> mp5; int main(){
n=read();
rin(i,1,n){
int a[6];
a[1]=read(),a[2]=read(),a[3]=read(),a[4]=read(),a[5]=read();
std::sort(a+1,a+6);
int x=a[1],y=a[2],z=a[3],v=a[4],w=a[5];
++mp1[x];
++mp1[y];
++mp1[z];
++mp1[v];
++mp1[w];
++mp2[mktw(x,y)];
++mp2[mktw(x,z)];
++mp2[mktw(x,v)];
++mp2[mktw(x,w)];
++mp2[mktw(y,z)];
++mp2[mktw(y,v)];
++mp2[mktw(y,w)];
++mp2[mktw(z,v)];
++mp2[mktw(z,w)];
++mp2[mktw(v,w)];
++mp3[mktr(x,y,z)];
++mp3[mktr(x,y,v)];
++mp3[mktr(x,y,w)];
++mp3[mktr(x,z,v)];
++mp3[mktr(x,z,w)];
++mp3[mktr(x,v,w)];
++mp3[mktr(y,z,v)];
++mp3[mktr(y,z,w)];
++mp3[mktr(y,v,w)];
++mp3[mktr(z,v,w)];
++mp4[mkfo(x,y,z,v)];
++mp4[mkfo(x,y,z,w)];
++mp4[mkfo(x,y,v,w)];
++mp4[mkfo(x,z,v,w)];
++mp4[mkfo(y,z,v,w)];
++mp5[mkfi(x,y,z,v,w)];
}
LL ans=0;
rin(i,1,1000000) ans+=1ll*mp1[i]*(mp1[i]-1)/2;
for(register std::map<tw,int>::iterator it=mp2.begin();it!=mp2.end();++it) ans-=1ll*it->second*(it->second-1)/2;
for(register std::map<tr,int>::iterator it=mp3.begin();it!=mp3.end();++it) ans+=1ll*it->second*(it->second-1)/2;
for(register std::map<fo,int>::iterator it=mp4.begin();it!=mp4.end();++it) ans-=1ll*it->second*(it->second-1)/2;
for(register std::map<fi,int>::iterator it=mp5.begin();it!=mp5.end();++it) ans+=1ll*it->second*(it->second-1)/2;
printf("%lld\n",1ll*n*(n-1)/2-ans);
return 0;
}

T3

#include <bits/stdc++.h>

#define rin(i,a,b) for(register int i=(a);i<=(b);++i)
#define irin(i,a,b) for(register int i=(a);i>=(b);--i)
#define trav(i,a) for(register int i=head[a];i;i=e[i].nxt)
typedef long long LL; using std::cin;
using std::cout;
using std::endl; inline int read(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
return x*f;
} const int MAXN=10005;
const int MAXK=1005; int n,k,a[MAXN],f[MAXN]; int main(){
n=read(),k=read();
rin(i,1,n) a[i]=read();
f[0]=0;
rin(i,1,n){
int temp=0;
irin(j,i,std::max(i-k+1,1)){
temp=std::max(temp,a[j]);
f[i]=std::max(f[i],f[j-1]+temp*(i-j+1));
}
}
printf("%d\n",f[n]);
return 0;
}

USACO2018DEC GOLD的更多相关文章

  1. Baskets of Gold Coins_暴力

    Problem Description You are given N baskets of gold coins. The baskets are numbered from 1 to N. In ...

  2. [UCSD白板题] Take as Much Gold as Possible

    Problem Introduction This problem is about implementing an algorithm for the knapsack without repeti ...

  3. XidianOJ 1120 Gold of Orz Pandas

    题目描述 Orz Panda is addicted to one RPG game. To make his character stronger, he have to fulfil tasks ...

  4. POJ 3274 Gold Balanced Lineup

    Gold Balanced Lineup Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10924 Accepted: 3244 ...

  5. 哈希-Gold Balanced Lineup 分类: POJ 哈希 2015-08-07 09:04 2人阅读 评论(0) 收藏

    Gold Balanced Lineup Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 13215 Accepted: 3873 ...

  6. Gold Coins 分类: POJ 2015-06-10 15:04 16人阅读 评论(0) 收藏

    Gold Coins Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 21767   Accepted: 13641 Desc ...

  7. 2012 #5 Gold miner

    Gold miner Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total ...

  8. Pots of gold game:看谁拿的钱多

    问题描述: Pots of gold game: Two players A & B. There are pots of gold arranged in a line, each cont ...

  9. [LOJ 1030] Discovering Gold

    B - Discovering Gold Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu ...

随机推荐

  1. [BZOJ 2002] [HNOI2010]弹飞绵羊(Link Cut Tree)

    [BZOJ 2002] [HNOI2010]弹飞绵羊(Link Cut Tree) 题面 某天,Lostmonkey发明了一种超级弹力装置,为了在他的绵羊朋友面前显摆,他邀请小绵羊一起玩个游戏.游戏一 ...

  2. Bat批处理文件入门

    这个东西吧,感觉在使用windows时作用不是很大,毕竟在windows环境下命令行确实用的比较少,但有时候也会用到,比如测试算法程序时使用批处理+文件可以省去每次手动输入.赶巧最近对批处理也比较感兴 ...

  3. jquery中的obj.attr()和obj.data

    实例一 obj.attr('data-max-width','aa'): obj.data('max-width') 问题 data只会获取第一次select赋值的值 区别 .data每次是从jque ...

  4. 25 Python之模块与包

    一.模块   模块就是一个包含了python定义和申明的文件,文件名就是模块的名字加上.py的后缀/ 模块的分类:     1.使用python编写的py文件     2.已被编译位共享库或者DLL或 ...

  5. vue路由守卫触发顺序

    不同组件之间的路由跳转流程图 导航被触发(A–>B) 调用A组件内路由守卫beforeRouteLeave(to,from,next) 调用全局路由前置守卫router.beforeEach(t ...

  6. 目录:Matrix Differential Calculus with Applications in Statistics and Econometrics,3rd_[Magnus2019]

    目录:Matrix Differential Calculus with Applications in Statistics and Econometrics,3rd_[Magnus2019] Ti ...

  7. linux创建定时任务发送钉钉通知

    一.现在钉钉里面添加机器人 添加成功后,复制出Webhook链接. 注意,自定义关键字时你的发送信息中一定要完整包含关键字 二.找到自己的服务器 1. sudo su 切换到root用户 2.cron ...

  8. Delphi 子界类型

  9. SSD源码解读——网络测试

    之前,对SSD的论文进行了解读,可以回顾之前的博客:https://www.cnblogs.com/dengshunge/p/11665929.html. 为了加深对SSD的理解,因此对SSD的源码进 ...

  10. 浅析Java泛型

    什么是泛型? 泛型是JDK 1.5的一项新特性,它的本质是参数化类型(Parameterized Type)的应用,也就是说所操作的数据类型被指定为一个参数,在用到的时候在指定具体的类型.这种参数类型 ...