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hdu 4022 Bombing (离散化)

·348 words·1 min
Note: This article is available in Chinese only. 本文暂无英文版本。 View original

wa了两次,原因是在同一个点可能有多个基地。。。

所以用set 是错误的,应该用multiset

然后因为这道题看到了map+set实现离散化的另外一种写法

我的代码:

代码实现
 1/*************************************************************************
 2	> File Name: code/hdoj/4022.cpp
 3	> Author: 111qqz
 4	> Email: rkz2013@126.com
 5	> Created Time: 2015年08月01日 星期六 04时37分20秒
 6 ************************************************************************/
 7
 8#include<iostream>
 9#include<iomanip>
10#include<cstdio>
11#include<algorithm>
12#include<cmath>
13#include<cstring>
14#include<string>
15#include<map>
16#include<set>
17#include<queue>
18#include<vector>
19#include<stack>
20#define y0 abc111qqz
21#define y1 hust111qqz
22#define yn hez111qqz
23#define j1 cute111qqz
24#define tm crazy111qqz
25#define lr dying111qqz
26using namespace std;
27#define REP(i, n) for (int i=0;i<int(n);++i)
28typedef long long LL;
29typedef unsigned long long ULL;
30const int N=2E5+7;
31const int inf = 0x7fffffff;
32
33map<int,int>xmap,ymap;
34multiset<int>x[N];
35multiset<int>y[N];
36int main()
37{
38    int n,m;
39    while (scanf("%d %d",&n,&m)!=EOF)
40    {
41	if (n==0&&m==0) break;
42	for ( int i = 1 ; i <= n ; i++)
43	{
44	    x[i].clear();
45	    y[i].clear();
46	}
47	xmap.clear();
48	ymap.clear();
49	int tx,ty;
50	int cntx=0,cnty=0;
51	for ( int i = 1 ; i <= n ; i++ )
52	{
53	    scanf("%d %d",&tx,&ty);
54	    if (!xmap[tx]) xmap[tx]=++cntx;
55	    if (!ymap[ty]) ymap[ty]=++cnty;
56	    x[xmap[tx]].insert(ymap[ty]);
57	    y[ymap[ty]].insert(xmap[tx]);
58	}
59	int c,d;
60	set<int>::iterator it;
61	for ( int i = 1; i <= m ; i++ )
62	{
63	    scanf("%d %d",&c,&d);
64	    if (c==0)
65	    {
66		cout<<x[xmap[d]].size()<<endl;
67		for ( it = x[xmap[d]].begin();it!=x[xmap[d]].end();it++)
68		{
69		    y[*it].erase(xmap[d]);
70		}
71		x[xmap[d]].clear();
72	    }
73	    else
74	    {
75		cout<<y[ymap[d]].size()<<endl;
76		for ( it =y[ymap[d]].begin();it!=y[ymap[d]].end();it++)
77		{
78		    x[*it].erase(ymap[d]);
79
80		}
81		y[ymap[d]].clear();
82	    }
83	}
84	printf("\n");
85    }
86
87	return 0;
88}

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简单模拟,n,m貌似给反了(两个地方给的不一致 ) 害我wa了两发 代码实现 1 2 /************************************************************************* 3 > File Name: code/2015summer/#5/K.cpp 4 > Author: 111qqz 5 > Email: rkz2013@126.com 6 > Created Time: 2015年07月30日 星期四 14时00分56秒 7 ************************************************************************/ 8 9 #include<iostream> 10 #include<iomanip> 11 #include<cstdio> 12 #include<algorithm> 13 #include<cmath> 14 #include<cstring> 15 #include<string> 16 #include<map> 17 #include<set> 18 #include<queue> 19 #include<vector> 20 #include<stack> 21 #define y0 abc111qqz 22 #define y1 hust111qqz 23 #define yn hez111qqz 24 #define j1 cute111qqz 25 #define tm crazy111qqz 26 #define lr dying111qqz 27 using namespace std; 28 #define REP(i, n) for (int i=0;i<int(n);++i) 29 typedef long long LL; 30 typedef unsigned long long ULL; 31 const int inf = 0x7fffffff; 32 const int N=1e2+5; 33 int b[N][N]; 34 int n,m; 35 char cmd[505]; 36 bool vis[N][N]; 37 int nx,ny; 38 int dx[4]={-1,0,1,0}; 39 int dy[4]={0,1,-0,-1}; 40 char ch[N][N]; 41 int main() 42 { 43 cin>>n>>m; 44 nx = 0; 45 ny = 0; 46 for ( int i = 0 ; i < n ; i++) 47 cin>>ch[i]; 48 for ( int i = 0 ; i <n ; i++ ) 49 { 50 for ( int j = 0 ; j < m ; j++ ) 51 { 52 b[i+1][j+1]=(int)(ch[i][j]-'0'); 53 } 54 } 55 int ans = 0; 56 memset(vis,false,sizeof(vis)); 57 int dir = 1; 58 cin>>cmd; 59 int len = strlen(cmd); 60 for ( int i = 0 ; i < len ; i ++ ) 61 { 62 // cout<<"ans:"<<ans<<endl; 63 // cout<<"nx:"<<nx<<" ny:"<<ny<<endl; 64 if (cmd[i]=='L') 65 { 66 dir = (dir+3)%4; 67 } 68 if (cmd[i]=='R') 69 { 70 dir = (dir+1)%4; 71 } 72 if (cmd[i]=='M') 73 { 74 if (dir==0) 75 { 76 if (vis[nx][ny]) 77 { 78 ans = ans + b[nx][ny]/2; 79 } 80 else 81 { 82 ans = ans + b[nx][ny]; 83 } 84 if (vis[nx][ny+1]) 85 { 86 ans = ans + b[nx][ny+1]/2; 87 } 88 else 89 { 90 ans = ans + b[nx][ny+1]; 91 } 92 vis[nx][ny]=true; 93 vis[nx][ny+1]=true; 94 nx = nx +dx[dir]; 95 ny = ny +dy[dir]; 96 } 97 if (dir==2) 98 { 99 nx = nx + dx[dir]; 100 ny = ny + dy[dir]; 101 if (vis[nx][ny]) 102 { 103 ans = ans + b[nx][ny]/2; 104 } 105 else 106 { 107 ans = ans + b[nx][ny]; 108 } 109 if (vis[nx][ny+1]) 110 { 111 ans = ans + b[nx][ny+1]/2; 112 } 113 else 114 { 115 ans = ans + b[nx][ny+1]; 116 } 117 vis[nx][ny]=true; 118 vis[nx][ny+1]=true; 119 } 120 if (dir==1) 121 { 122 nx = nx + dx[dir]; 123 ny = ny + dy[dir]; 124 if (vis[nx][ny]) 125 { 126 ans = ans + b[nx][ny]/2; 127 } 128 else 129 { 130 ans = ans + b[nx][ny]; 131 } 132 if (vis[nx+1][ny]) 133 { 134 ans = ans + b[nx+1][ny]/2; 135 } 136 else 137 { 138 ans = ans + b[nx+1][ny]; 139 } 140 vis[nx][ny]=true; 141 vis[nx+1][ny]=true; 142 143 } 144 if (dir==3) 145 { 146 147 if (vis[nx][ny]) 148 { 149 ans = ans + b[nx][ny]/2; 150 } 151 else 152 { 153 ans = ans + b[nx][ny]; 154 } 155 if (vis[nx+1][ny]) 156 { 157 ans = ans + b[nx+1][ny]/2; 158 } 159 else 160 { 161 ans = ans + b[nx+1][ny]; 162 } 163 vis[nx][ny]=true; 164 vis[nx+1][ny]=true; 165 nx = nx +dx[dir]; 166 ny = ny +dy[dir]; 167 } 168 } 169 170 } 171 cout<<ans<<endl; 172 173 return 0; 174 }