Escape

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 10382 Accepted Submission(s): 2485

Problem Description

2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.

Input

More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000

Output

Determine whether all people can live up to these stars
If you can output YES, otherwise output NO.

Sample Input

1 1

2 2

1 0

1 1

Sample Output

YES

Source

2010 ACM-ICPC Multi-University Training Contest（17）——Host by ZSTU

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lcy

题意：

　　给你n个人m个星球，和第i个人能否适应第j个星球，1为适应，0为不适应。问你全部人能不能去星球上。

　　矩阵建边，跑一下二分图多重匹配。如果这个人无法去任意星球，直接break。

普通版：1560ms

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <string>
 5 #include <algorithm>
 6 #include <cmath>
 7 #include <vector>
 8 #include <queue>
 9 #include <map>
10 #include <stack>
11 #include <set>
12 using namespace std;
13 typedef long long LL;
14 #define ms(a, b) memset(a, b, sizeof(a))
15 #define pb push_back
16 #define mp make_pair
17 const LL INF = 0x7fffffff;
18 const int inf = 0x3f3f3f3f;
19 const int mod = 1e9+7;
20 const int maxn = 100000+10;
21 const int maxm = 10+10;
22 int n, m, uN, vN;
23 int g[maxn][maxm];
24 int linker[maxm][maxn];
25 bool used[maxm];
26 int num[maxm];
27 bool dfs(int u)
28 {
29     for(int v = 0;v<vN;v++)
30         if(g[u][v] && !used[v]){
31             used[v] = true;
32             if(linker[v][0]<num[v]){
33                 linker[v][++linker[v][0]] = u;
34                 return true;
35             }
36             for(int i = 1;i<=num[0];i++)
37                 if(dfs(linker[v][i])){
38                     linker[v][i] = u;
39                     return true;
40                 }
41         }
42     return false;
43 }
44 int hungary()
45 {
46     int res = 0;
47     for(int i = 0;i<vN;i++){
48         linker[i][0] = 0;
49     }
50     for(int u = 0;u<uN;u++){
51         ms(used, false);
52         if(dfs(u))  res++;
53         else return res;
54     }
55     return res;
56 }
57 void init() {
58     ms(g, 0);
59 }
60 void solve() {
61     for(int i = 0;i<n;i++){
62         for(int j = 0;j<m;j++){
63             int x;
64             scanf("%d", &x);
65             if(x==1){
66                 g[i][j] = 1;
67             }
68             else{
69                 g[i][j] = 0;
70             }
71         }
72     }
73     for(int i = 0;i<m;i++)
74         scanf("%d", &num[i]);
75     vN = m, uN = n;
76     int ans = hungary();
77 //    printf("%d\n", ans);
78     if(ans==n){
79         printf("YES\n");
80     }
81     else{
82         printf("NO\n");
83     }
84 }
85 int main() {
86 #ifdef LOCAL
87     freopen("input.txt", "r", stdin);
88 //        freopen("output.txt", "w", stdout);
89 #endif
90     while(~scanf("%d%d", &n, &m)){
91         init();
92         solve();
93     }
94     return 0;
95 }

View Code

fread版：249ms

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <string>
  5 #include <algorithm>
  6 #include <cmath>
  7 #include <vector>
  8 #include <queue>
  9 #include <map>
 10 #include <stack>
 11 #include <set>
 12 using namespace std;
 13 typedef long long LL;
 14 #define ms(a, b) memset(a, b, sizeof(a))
 15 #define pb push_back
 16 #define mp make_pair
 17 const LL INF = 0x7fffffff;
 18 const int inf = 0x3f3f3f3f;
 19 const int mod = 1e9+7;
 20 const int maxn = 100000+10;
 21 const int maxm = 10+10;
 22 //输入挂
 23 const int MAXBUF = 10000;
 24 char buf[MAXBUF], *ps = buf, *pe = buf+1;
 25 inline void rnext()
 26 {
 27     if(++ps == pe)
 28         pe = (ps = buf)+fread(buf,sizeof(char),sizeof(buf)/sizeof(char),stdin);
 29 }
 30 template <class T>
 31 inline bool in(T &ans)
 32 {
 33     ans = 0;
 34     T f = 1;
 35     if(ps == pe) return false;//EOF
 36     do{
 37         rnext();
 38         if(‘-‘ == *ps) f = -1;
 39     }while(!isdigit(*ps) && ps != pe);
 40     if(ps == pe) return false;//EOF
 41     do
 42     {
 43         ans = (ans<<1)+(ans<<3)+*ps-48;
 44         rnext();
 45     }while(isdigit(*ps) && ps != pe);
 46     ans *= f;
 47     return true;
 48 }
 49 const int  MAXOUT=10000;
 50 char bufout[MAXOUT], outtmp[50],*pout = bufout, *pend = bufout+MAXOUT;
 51 inline void write()
 52 {
 53     fwrite(bufout,sizeof(char),pout-bufout,stdout);
 54     pout = bufout;
 55 }
 56 inline void out_char(char c){ *(pout++) = c;if(pout == pend) write();}
 57 inline void out_str(char *s)
 58 {
 59     while(*s)
 60     {
 61         *(pout++) = *(s++);
 62         if(pout == pend) write();
 63     }
 64 }
 65 template <class T>
 66 inline void out_int(T x) {
 67     if(!x)
 68     {
 69         out_char(‘0‘);
 70         return;
 71     }
 72     if(x < 0) x = -x,out_char(‘-‘);
 73     int len = 0;
 74     while(x)
 75     {
 76         outtmp[len++] = x%10+48;
 77         x /= 10;
 78     }
 79     outtmp[len] = 0;
 80     for(int i = 0, j = len-1; i < j; i++,j--) swap(outtmp[i],outtmp[j]);
 81     out_str(outtmp);
 82 }
 83 //end
 84 int n, m, uN, vN;
 85 int g[maxn][maxm];
 86 int linker[maxm][maxn];
 87 bool used[maxm];
 88 int num[maxm];
 89 bool dfs(int u)
 90 {
 91     for(int v = 0;v<vN;v++)
 92         if(g[u][v] && !used[v]){
 93             used[v] = true;
 94             if(linker[v][0]<num[v]){
 95                 linker[v][++linker[v][0]] = u;
 96                 return true;
 97             }
 98             for(int i = 1;i<=num[0];i++)
 99                 if(dfs(linker[v][i])){
100                     linker[v][i] = u;
101                     return true;
102                 }
103         }
104     return false;
105 }
106 int hungary()
107 {
108     int res = 0;
109     for(int i = 0;i<vN;i++){
110         linker[i][0] = 0;
111     }
112     for(int u = 0;u<uN;u++){
113         ms(used, false);
114         if(dfs(u))  res++;
115         else return res;
116     }
117     return res;
118 }
119 void init() {
120     ms(g, 0);
121 }
122 void solve() {
123     int x;
124     for(int i = 0;i<n;i++){
125         for(int j = 0;j<m;j++){
126             in(x);
127             if(x==1){
128                 g[i][j] = 1;
129             }
130             else{
131                 g[i][j] = 0;
132             }
133         }
134     }
135     for(int i = 0;i<m;i++)
136         in(num[i]);
137     vN = m, uN = n;
138     int ans = hungary();
139     if(ans==n){
140         out_str("YES");out_char(‘\n‘);
141     }
142     else{
143         out_str("NO");out_char(‘\n‘);
144     }
145 }
146 int main() {
147 #ifdef LOCAL
148     freopen("input.txt", "r", stdin);
149 //        freopen("output.txt", "w", stdout);
150 #endif
151     while(in(n)&&in(m)){
152         init();
153         solve();
154     }
155     write();
156     return 0;
157 }

View Code

HDU 3605 Escape（二分图多重匹配问题）

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HDU 3605 Escape（二分图多重匹配问题）

Escape

题意：

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