D - 粉碎叛乱F - 其他起义

D - 粉碎叛乱

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Description

Adam and Eve play a card game using a regular deck of 52 cards. The rules are simple. The players sit on opposite sides of a table, facing each other. Each player gets k cards from the deck and, after looking at them, places the cards face down in a row on the table. Adam’s cards are numbered from 1 to k from his left, and Eve’s cards are numbered 1 to k from her right (so Eve’s i:th card is opposite Adam’s i:th card). The cards are turned face up, and points are awarded as follows (for each i ∈ {1, . . . , k}):

If Adam’s i:th card beats Eve’s i:th card, then Adam gets one point.

If Eve’s i:th card beats Adam’s i:th card, then Eve gets one point.

A card with higher value always beats a card with a
lower value: a three beats a two, a four beats a three and a two, etc.
An ace beats every card except (possibly) another ace.

If the two i:th cards have the same value, then the
suit determines who wins: hearts beats all other suits, spades beats all
suits except hearts, diamond beats only clubs, and clubs does not beat
any suit.

For example, the ten of spades beats the ten of diamonds but not the Jack of clubs.

This ought to be a game of chance, but lately Eve is
winning most of the time, and the reason is that she has started to use
marked cards. In other words, she knows which cards Adam has on the
table before he turns them face up. Using this information she orders
her own cards so that she gets as many points as possible.

Your task is to, given Adam’s and Eve’s cards, determine how many points Eve will get if she plays optimally.

Input

There will be several test cases. The first line of input will contain a
single positive integer N giving the number of test cases. After that
line follow the test cases.

Each test case starts with a line with a single positive
integer k <= 26 which is the number of cards each player gets. The
next line describes the k cards Adam has placed on the table, left to
right. The next line describes the k cards Eve has (but she has not yet
placed them on the table). A card is described by two characters, the
first one being its value (2, 3, 4, 5, 6, 7, 8 ,9, T, J, Q, K, or A),
and the second one being its suit (C, D, S, or H). Cards are separated
by white spaces. So if Adam’s cards are the ten of clubs, the two of
hearts, and the Jack of diamonds, that could be described by the line

TC 2H JD

Output

For each test case output a single line with the number of points Eve
gets if she picks the optimal way to arrange her cards on the table.

Sample Input

3
1
JD
JH
2
5D TC
4C 5H
3
2H 3H 4H
2D 3D 4D

Sample Output

1 1 2

题解：坑，A竟然最大。。

代码：

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<cmath>

 #include<algorithm>

 #define mem(x,y) memset(x,y,sizeof(x))

 using namespace std;

 typedef long long LL;

 const int INF=0x3f3f3f3f;

 int a[],b[];

 char s[];

 int getnum(){

     int x=;

     if(s[]=='A')x=;

     else if(s[]=='T')x=;

     else if(s[]=='J')x=;

     else if(s[]=='Q')x=;

     else if(s[]=='K')x=;

     else if(s[]>=''&&s[]<='')x=s[]-'';

     x=x*;

     if(s[]=='C')x+=;

     else if(s[]=='D')x+=;

     else if(s[]=='S')x+=;

     else if(s[]=='H')x+=;

      return x;

 }

 int main(){

     int T,N;

     scanf("%d",&T);

     while(T--){

         scanf("%d",&N);

         for(int i=;i<N;i++){

             scanf("%s",s);

             a[i]=getnum();

         }

         for(int i=;i<N;i++){

             scanf("%s",s);

             b[i]=getnum();

         }

     //    for(int i=0;i<N;i++)printf("%d ",a[i]);puts("");

     //    for(int i=0;i<N;i++)printf("%d ",b[i]);puts("");

         sort(a,a+N);sort(b,b+N);

         int ans=;

         for(int i=,j=;i<N&&j<N;){

             if(b[i]>=a[j]){

                 i++;j++;ans++;

             }

             else i++;

         }

         printf("%d\n",ans);

     }

     return ;

 }

F - 其他起义

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 1610

Description

Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.

Your task is counting the segments of different colors you can see at last.

Input

The first line of each data set contains exactly one integer n, 1 <= n <= 8000, equal to the number of colored segments.

Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces:

x1 x2 c

x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates the color of the segment.

All the numbers are in the range [0, 8000], and they are all integers.

Input may contain several data set, process to the end of file.

Output

Each line of the output should contain a color index that can be seen
from the top, following the count of the segments of this color, they
should be printed according to the color index.

If some color can't be seen, you shouldn't print it.

Print a blank line after every dataset.

Sample Input

5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3
4
0 1 1
3 4 1
1 3 2
1 3 1
6
0 1 0
1 2 1
2 3 1
1 2 0
2 3 0
1 2 1

Sample Output

1 1
2 1
3 1

1 1

0 2
1 1

题解：暴力；

代码：

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

#define mem(x,y) memset(x,y,sizeof(x))

using namespace std;

typedef long long LL;

const int INF=0x3f3f3f3f;

const int MAXN=1e6+;

int tree[MAXN],pre[MAXN];

int main(){

    int N,x,y,z;

    while(~scanf("%d",&N)){

        mem(tree,-);mem(pre,);

        for(int i=;i<N;i++){

            scanf("%d%d%d",&x,&y,&z);

            for(int j=x;j<y;j++)tree[j]=z;

        }

        for(int i=;i<=;i++){

            if(tree[i]==-)continue;

                pre[tree[i]]++;

            while(tree[i+]==tree[i])i++;

        }

        for(int i=;i<=;i++){

            if(!pre[i])continue;

            printf("%d %d\n",i,pre[i]);

        }

        puts("");

    }

    return ;

}

巴特西

D - 粉碎叛乱F - 其他起义

最新文章

热门文章

巴特西

﻿ D - 粉碎叛乱F - 其他起义

最新文章

热门文章

D - 粉碎叛乱F - 其他起义