题目描述
随着海上运输石油泄漏的问题，一个新的有利可图的行业正在诞生，那就是撇油行业。如今，在墨西哥湾漂浮的大量石油，吸引了许多商人的目光。这些商人们有一种特殊的飞机，可以一瓢略过整个海面20米乘10米这么大的长方形。（上下相邻或者左右相邻的格子，不能斜着来）当然，这要求一瓢撇过去的全部是油，如果一瓢里面有油有水的话，那就毫无意义了，资源完全无法利用。现在，商人想要知道，在这片区域中，他可以最多得到多少瓢油。
地图是一个N×N的网络，每个格子表示10m×10m的正方形区域，每个区域都被标示上了是油还是水
输入描述:
测试输入包含多条测试数据
测试数据的第一行给出了测试数据的数目T（T<75）
每个测试样例都用数字N（N<50）来表示地图区域的大小，接下来N行，每行都有N个字符，其中符号’.’表示海面、符号’#’表示油面。
输出描述:
输出格式如下“Case X: M”（X从1开始），M是商人可以最多得到的油量。
示例1
输入
1 6 ...... .##... ...... .#..#. .#..## ......
输出
Case 1: 3
1 #include <iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<stdlib.h>
5 #include<limits.h>
6 #include<queue>
7 #include<cmath>
8 #include<algorithm>
9 using namespace std;
10 #define maxn 2505
11 char str[maxn][maxn];
12 int ans[maxn];
13 int map[maxn][maxn];
14 int a[maxn], b[maxn];
15 int n;
16 int v[maxn];
17 int dfs(int x)
18 {
19 for (int i = 1; i <= n * n; i++)
20 {
21 if (map[x][i] && !ans[i])
22 {
23 ans[i] = 1;
24 if (!v[i] || dfs(v[i]))
25 {
26 v[i] = x;
27 return 1;
28 }
29 }
30 }
31 return 0;
32 }
33 int main()
34 {
35 int i, j, T, cases = 0;
36 scanf_s("%d", &T);
37 while (T--)
38 {
39 int x;
40 scanf_s("%d", &n);
41 memset(v, 0, sizeof(v));
42 memset(map, 0, sizeof(map));
43 memset(str, '\0', sizeof(str));
44 for (i = 1; i <= n; i++)
45 for (j = 1; j <= n; j++)
46 cin >> str[i][j];
47 for (i = 1; i <= n; i++)
48 for (j = 1; j <= n; j++)
49 {
50 if (str[i][j] == '#' && str[i][j - 1] == '#')
51 map[(i - 1)*n + j][(i - 1)*n + j - 1] = 1;
52 if (str[i][j] == '#' && str[i][j + 1] == '#')
53 map[(i - 1)*n + j][(i - 1)*n + j + 1] = 1;
54 if (str[i][j] == '#' && str[i - 1][j] == '#')
55 map[(i - 1)*n + j][(i - 2)*n + j] = 1;
56 if (str[i][j] == '#' && str[i + 1][j] == '#')
57 map[(i - 1)*n + j][(i)*n + j] = 1;
58 }
59 /*for (i = 1; i <= n * n; i++)
60 {
61 for (j = 1; j <= n * n; j++)
62 {
63 cout << map[i][j];
64 }
65 cout << endl;
66 }*/
67 int sum = 0;
68 for (i = 1; i <= n * n; i++)
69 {
70 memset(ans, 0, sizeof(ans));
71 if (dfs(i)) sum++;
72 }
73 printf("Case %d: %d\n", ++cases, sum / 2);
74 }
75 return 0;
76 }
居然可以用二分匹配
1 #include <iostream>
2 #include <queue>
3 #include <algorithm>
4 #include <string.h>
5 #include <math.h>
6 #include <map>
7 using namespace std;
8 const int maxn=2000;
9 int girl[maxn],used[maxn],line[maxn][maxn],path[60][60],temp1,temp2;
10 char a[60][60];
11 bool find(int x)
12 {
13 for (int i=1;i<temp2;i++)
14 if (line[x][i]&&!used[i]) //x与i有关系
15 {
16 used[i]=1;
17 if (girl[i]==0||find(girl[i])) //名花无主或者还能腾位置；
18 {
19 girl[i]=x;
20 return true;
21 }
22 }
23 return false;
24 }
25 int main()
26 {
27
28 int t,n,ans,tt;
29 cin>>t;
30 tt=1;
31 while (t--)
32 {
33 ans=0;
34 temp1=temp2=1;
35 memset(line,0,sizeof(line));
36 memset(girl,0,sizeof(girl));
37 memset(path,0,sizeof(path));
38 cin>>n;
39 for (int i=1;i<=n;i++)
40 for (int j=1;j<=n;j++)
41 {
42 cin>>a[i][j];
43 if ((i+j)%2==0)
44 path[i][j]=temp1++;
45 else
46 path[i][j]=temp2++;
47 }
48 for (int i=1;i<=n;i++)
49 for (int j=1;j<=n;j++)
50 {
51 if ((i+j)%2==1&&a[i][j]=='#')
52 {
53 if (path[i-1][j]>=1&&a[i-1][j]=='#')
54 line[path[i-1][j]][path[i][j]]=1;
55 if (path[i+1][j]>=1&&a[i+1][j]=='#')
56 line[path[i+1][j]][path[i][j]]=1;
57 if (path[i][j-1]>=1&&a[i][j-1]=='#')
58 line[path[i][j-1]][path[i][j]]=1;
59 if (path[i][j+1]>=1&&a[i][j+1]=='#')
60 line[path[i][j+1]][path[i][j]]=1;
61 }
62 }
63 for (int i=1;i<temp1;i++)
64 {
65 memset(used,0,sizeof(used));
66 if (find(i))
67 ans++;
68 }
69 cout<<"Case "<<tt++<<": "<<ans<<endl;
70 }
71 }
dfs也可以做
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 using namespace std;
5 int n;
6 char s[55][55];
7 int num1,num2;
8 void dfs(int x,int y)
9 {
10 if(x>=1&&x<=n&&y>=1&&y<=n&&s[x][y]=='#')
11 {
12 if((x+y)%2) num1++;
13 else num2++;
14 s[x][y]='.';
15 dfs(x,y-1);
16 dfs(x,y+1);
17 dfs(x-1,y);
18 dfs(x+1,y);
19 }
20 }
21 int main()
22 {
23 int T;
24 scanf("%d",&T);
25 int CASE=1;
26 while(T--)
27 {
28 scanf("%d",&n);
29 int i,j;
30 for(i=1; i<=n; i++)
31 {
32 getchar();
33 for(j=1; j<=n; j++)
34 scanf("%c",&s[i][j]);
35 }
36 int ans=0;
37 for(i=1; i<=n; i++)
38 {
39 for(j=1; j<=n; j++)
40 {
41 if(s[i][j]=='#')
42 {
43 num1=0,num2=0;
44 dfs(i,j);
45 ans+=min(num1,num2);
46 }
47 }
48 }
49 printf("Case %d: %d\n",CASE++,ans);
50 }
51 }
1 #include <cstdio>
2 #include <iostream>
3 #include <cstring>
4 #include <map>
5 #include <set>
6 #include <bitset>
7 #include <cctype>
8 #include <cstdlib>
9 #include <queue>
10 #include <cmath>
11 #include <stack>
12 #include <ctime>
13 #include <string>
14 #include <vector>
15 #include <sstream>
16 #include <functional>
17 #include <algorithm>
18 using namespace std;
19
20 #define mem(a,n) memset(a,n,sizeof(a))
21 #define memc(a,b) memcpy(a,b,sizeof(b))
22 #define rep(i,a,n) for(int i=a;i<n;i++) ///[a,n)
23 #define dec(i,n,a) for(int i=n;i>=a;i--)///[n,a]
24 #define pb push_back
25 #define fi first
26 #define se second
27 #define IO ios::sync_with_stdio(false)
28 #define fre freopen("in.txt","r",stdin)
29 #define lson l,m,rt<<1
30 #define rson m+1,r,rt<<1|1
31 typedef long long ll;
32 typedef unsigned long long ull;
33 const double PI=acos(-1.0);
34 const double E=2.718281828459045;
35 const double eps=1e-8;
36 const int INF=0x3f3f3f3f;
37 const int MOD=258280327;
38 const int N=50+5;
39 const ll maxn=1e6+5;
40 const int dir[4][2]= {-1,0,1,0,0,-1,0,1};
41 int n,ans;
42 char a[N][N];
43 bool vis[N][N];
44 bool in(int x,int y)
45 {
46 if(x>=0&&x<n&&y>=0&&y<n) return true;
47 return false;
48 }
49 bool dfs(int x,int y)
50 {
51 rep(i,0,4)
52 {
53 int nx=x+dir[i][0],ny=y+dir[i][1];
54 if(in(nx,ny) && a[nx][ny]=='#')
55 {
56 a[x][y]='.';
57 if(!dfs(nx,ny))
58 {
59 a[nx][ny]='#';
60 ans++;
61 return true;
62 }
63 a[x][y]='#';
64 }
65 }
66 return false;
67 }
68 int main()
69 {
70 int T;
71 scanf("%d",&T);
72 rep(cas,0,T)
73 {
74 scanf("%d",&n);
75 ans=0;
76 mem(vis,0);
77 rep(i,0,n) scanf("%s",a[i]);
78 rep(i,0,n)
79 {
80 rep(j,0,n)
81 {
82 if(a[i][j]=='#')
83 dfs(i,j);
84 }
85 }
86 printf("Case %d: %d\n",cas+1,ans);
87 }
88 return 0;
89 }
View Code
#include <cstdio>
#include <iostream>
#include <cstring>
#include <map>
#include <set>
#include <bitset>
#include <cctype>
#include <cstdlib>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <string>
#include <vector>
#include <sstream>
#include <functional>
#include <algorithm>
using namespace std;
#define mem(a,n) memset(a,n,sizeof(a))
#define memc(a,b) memcpy(a,b,sizeof(b))
#define rep(i,a,n) for(int i=a;i<n;i++) ///[a,n)
#define dec(i,n,a) for(int i=n;i>=a;i--)///[n,a]
#define pb push_back
#define fi first
#define se second
#define IO ios::sync_with_stdio(false)
#define fre freopen("in.txt","r",stdin)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long ll;
typedef unsigned long long ull;
const double PI=acos(-1.0);
const double E=2.718281828459045;
const double eps=1e-8;
const int INF=0x3f3f3f3f;
const int MOD=258280327;
const int N=50+5;
const ll maxn=1e6+5;
const int dir[4][2]= {-1,0,1,0,0,-1,0,1};
int n,ans;
char a[N][N];
bool vis[N][N];
bool in(int x,int y)
{
if(x>=0&&x<n&&y>=0&&y<n) return true;
return false;
}
bool dfs(int x,int y)
{
rep(i,0,4)
{
int nx=x+dir[i][0],ny=y+dir[i][1];
if(in(nx,ny) && a[nx][ny]=='#')
{
a[x][y]='.';
if(!dfs(nx,ny))
{
a[nx][ny]='#';
ans++;
return true;
}
a[x][y]='#';
}
}
return false;
}
int main()
{
int T;
scanf("%d",&T);
rep(cas,0,T)
{
scanf("%d",&n);
ans=0;
mem(vis,0);
rep(i,0,n) scanf("%s",a[i]);
rep(i,0,n)
{
rep(j,0,n)
{
if(a[i][j]=='#')
dfs(i,j);
}
}
printf("Case %d: %d\n",cas+1,ans);
}
return 0;
}
