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  • 15.3.17周练

    连接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=72057#overview

    密码acmore

    A：最小费用流

    结题报告：http://www.cnblogs.com/gj-Acit/p/4374951.html

      1 #include <map>
  2 #include <set>
  3 #include <stack>
  4 #include <queue>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <vector>
  8 #include <cstdio>
  9 #include <cctype>
 10 #include <cstring>
 11 #include <cstdlib>
 12 #include <iostream>
 13 #include <algorithm>
 14 using namespace std;
 15 #define INF 0x3f3f3f3f
 16 #define inf (-((LL)1<<40))
 17 #define lson k<<1, L, mid
 18 #define rson k<<1|1, mid+1, R
 19 #define mem0(a) memset(a,0,sizeof(a))
 20 #define mem1(a) memset(a,-1,sizeof(a))
 21 #define mem(a, b) memset(a, b, sizeof(a))
 22 #define FIN freopen("in.txt", "r", stdin)
 23 #define FOUT freopen("out.txt", "w", stdout)
 24 #define rep(i, a, b) for(int i = a; i <= b; i ++)
 25 
 26 template<class T> T CMP_MIN(T a, T b) { return a < b; }
 27 template<class T> T CMP_MAX(T a, T b) { return a > b; }
 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
 32 
 33 //typedef __int64 LL;
 34 typedef long long LL;
 35 const int MAXN = 10010;
 36 const int MAXM = 20010;
 37 const double eps = 1e-4;
 38 
 39 
 40 const int dx[4] = {0, 1, 0, -1};
 41 const int dy[4] = {1, 0, -1, 0};
 42 int T, N, M, K, B, ans = 0;
 43 
 44 /*******************************************************************/
 45 struct Edge {
 46     int to, cap, flow, cost, next;
 47     Edge(){}
 48     Edge(int _n, int _v, int _c, int _f, int _cost){
 49         next = _n; to = _v; cap = _c;
 50         flow = _f; cost = _cost;
 51     }
 52 };
 53 
 54 struct MCMF
 55 {
 56     int n, m, src, des;
 57     int head[MAXN], tot;
 58     Edge edges[MAXM];
 59     int inq[MAXN];
 60     int d[MAXN];
 61     int p[MAXN];
 62     int a[MAXN];
 63 
 64     void init(int n) {
 65         this->tot = 0;
 66         this->n = n;
 67         mem1(head);
 68     }
 69 
 70     void add_edge(int from, int to, int cap, int cost) {
 71         edges[tot] = Edge(head[from], to, cap, 0, cost);
 72         head[from] = tot ++;
 73         edges[tot] = Edge(head[to], from, 0, 0, -cost);
 74         head[to] = tot ++;
 75     }
 76 
 77     bool bellman_ford(int s, int t, int& flow) {
 78         for(int i = 0; i < n; i ++) {
 79             d[i] = INF;
 80             inq[i] = 0;
 81         }
 82         d[s] = 0; inq[s] = 1;
 83         p[s] = 0; a[s] = INF;
 84 
 85         queue<int>Q;
 86         Q.push(s);
 87         while(!Q.empty()) {
 88             int u = Q.front(); Q.pop();
 89             inq[u] = false;
 90             for(int i = head[u]; i != -1; i = edges[i].next) {
 91                 int v = edges[i].to;
 92                 if(edges[i].cap > edges[i].flow && d[v] > d[u] + edges[i].cost) {
 93                     d[v] = d[u] + edges[i].cost;
 94                     p[v] = i;
 95                     a[v] = min(a[u], edges[i].cap - edges[i].flow);
 96                     if(!inq[v]) {
 97                         Q.push(v);
 98                         inq[v] = 1;
 99                     }
100                 }
101             }
102         }
103         if(d[t] >= 0) return false;
104 
105         flow += a[t];
106         ans += d[t] * a[t];
107 
108         int u = t;
109         while(u != s) {
110             edges[p[u]].flow += a[t];
111             edges[p[u]^1].flow -= a[t];
112             u = edges[p[u]^1].to;
113         }
114         return true;
115     }
116 
117     int min_cost(int s, int t) {
118         int flow = 0;
119         while(bellman_ford(s, t, flow));
120         return ans;
121     }
122 
123 };
124 /***************************************************************/
125 
126 int idx(int i, int j) {
127     return (i - 1) * M + j;
128 }
129 
130 int ok(int i, int j) {
131     return (i >= 1 && i <= N && j >= 1 && j <= M);
132 }
133 
134 
135 MCMF mcmf;
136 
137 int main()
138 {
139     //FIN;
140     cin >> T;
141     rep (t, 1, T) {
142         scanf("%d %d %d", &N, &M, &K);
143         mcmf.init(N * M + 2);
144         mcmf.src = 0; mcmf.des = N * M + 1; ans = 0;
145         rep (i, 1, N) rep (j, 1, M) {
146             scanf("%d", &B);
147             ans += B * B;
148             if(i % 2 == j % 2) rep (k, 1, K) {
149                 mcmf.add_edge(mcmf.src, idx(i, j), 1, 2 * k - 1 - 2 * B);
150                 rep (k, 0, 3) if(ok(i + dx[k], j + dy[k])) {
151                     mcmf.add_edge(idx(i, j), idx(i + dx[k], j + dy[k]), INF, 0);
152                 }
153             }
154             else rep (k, 1, K) {
155                 mcmf.add_edge(idx(i, j), mcmf.des, 1, 2 * k - 1 - 2 * B);
156             }
157         }
158         printf("Case %d: %d
", t, mcmf.min_cost(mcmf.src, mcmf.des));
159     }
160     return 0;
161 }
    View Code

    B:贪心

    算出每只文字穿过球体的时间段，然后就是求所有时间段（线段）的最小覆盖次数，贪心

      1 #include <map>
  2 #include <set>
  3 #include <stack>
  4 #include <queue>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <vector>
  8 #include <cstdio>
  9 #include <cctype>
 10 #include <cstring>
 11 #include <cstdlib>
 12 #include <iostream>
 13 #include <algorithm>
 14 using namespace std;
 15 #define INF 0x3f3f3f3f
 16 #define inf (-((LL)1<<40))
 17 #define lson k<<1, L, mid
 18 #define rson k<<1|1, mid+1, R
 19 #define mem0(a) memset(a,0,sizeof(a))
 20 #define mem1(a) memset(a,-1,sizeof(a))
 21 #define mem(a, b) memset(a, b, sizeof(a))
 22 #define FIN freopen("in.txt", "r", stdin)
 23 #define FOUT freopen("out.txt", "w", stdout)
 24 #define rep(i, a, b) for(int i = a; i <= b; i ++)
 25 
 26 template<class T> T CMP_MIN(T a, T b) { return a < b; }
 27 template<class T> T CMP_MAX(T a, T b) { return a > b; }
 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
 32 
 33 //typedef __int64 LL;
 34 typedef long long LL;
 35 const int MAXN = 1100;
 36 const int MAXM = 2000010;
 37 const double eps = 1e-4;
 38 
 39 struct Point {
 40     double x, y, z;
 41     Point(){}
 42     Point(double _x, double _y, double _z):x(_x), y(_y), z(_z){}
 43 }O(0, 0, 0);
 44 struct Line{
 45     double a, b;
 46     bool operator < (const Line& A) const {
 47         return b != A.b ? b < A.b : a < A.a;
 48     }
 49 }l[110000];
 50 
 51 int n, T, cntL = 0, A, B;
 52 double R;
 53 
 54 double dis(Point a, Point b) {
 55     return (double)sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y) + (a.z - b.z)*(a.z - b.z));
 56 }
 57 
 58 int convert(Point p, Point v)
 59 {
 60     double a = v.x*v.x + v.y*v.y + v.z*v.z;
 61     double b = 2.0 * (p.x*v.x + p.y*v.y + p.z*v.z);
 62     double c = p.x*p.x + p.y*p.y + p.z*p.z - R * R;
 63     double del = b * b - 4.0 * a * c;
 64     if(del < 0) return 0;
 65     double x1 = (-b - sqrt(del)) / (2 * a), x2 = (-b + sqrt(del)) / (2 * a);
 66     if(x2 < 0) return 0;
 67     l[cntL].a = max(x1, 0.0);
 68     l[cntL++].b = x2;
 69     return 1;
 70 }
 71 
 72 int calcAnsB()
 73 {
 74     sort(l, l + cntL);
 75     double lasR = -1e10;
 76     int ret = 0;
 77     for(int i = 0; i < cntL; i ++) {
 78         if(l[i].a - lasR > eps) {
 79             ret ++;
 80             lasR = l[i].b;
 81         }
 82     }
 83     return ret;
 84 }
 85 
 86 int main()
 87 {
 88     //FIN;
 89     Point p, v;
 90     cin>>T;
 91     rep (t, 1, T) {
 92         scanf("%d %lf", &n, &R);
 93         A = 0; B = 0; cntL = 0;
 94         int x,y,z,a,b,c;
 95         rep (i, 0, n - 1) {
 96             scanf("%d %d %d %d %d %d", &x, &y, &z, &a, &b, &c);
 97             l[i].a = l[i].b = -1;
 98             p.x = x; p.y = y; p.z = z;
 99             v.x = a; v.y = b; v.z = c;
100             A += convert(p, v);
101         }
102 
103         B = calcAnsB();
104         printf("Case %d: %d %d
", t, A, B);
105     }
106     return 0;
107 }
    View Code

    C:Easy Game，Easy

     1 import java.util.Scanner;
 2 
 3 
 4 public class Main {
 5     public static void main(String[] args) {
 6         Scanner cin  = new Scanner(System.in);
 7         int t = cin.nextInt();
 8         for(int i = 1; i <= t; i ++) {
 9             String s = cin.next();
10             System.out.print("Case " + i + ": ");
11             if(s.length() % 2 == 1)System.out.println("Odd");
12             else System.out.println("Even");
13         }
14     }
    View Code

    D:贪心，每次取最接近x/2的数

     1 import java.math.BigInteger;
 2 import java.util.Scanner;
 3 
 4 
 5 public class Main {
 6     public static void main(String[] args) {
 7         Scanner cin  = new Scanner(System.in);
 8         BigInteger TWO = new BigInteger("2");
 9         int t = cin.nextInt();
10         for(int i = 1; i <= t; i ++) {
11             BigInteger A = cin.nextBigInteger();
12             BigInteger B = cin.nextBigInteger();
13             int ans = 0;
14             while(A.compareTo(B) > 0) {
15                 A = A.subtract(A.mod(A.divide(TWO).add(BigInteger.ONE)));
16                 ans ++;
17             }
18             System.out.println("Case " + i + ": " + ans);
19         }
20     }
21 }
    View Code

    E:求凸四边形个数，暴力枚举，若两条线断相交则这四个点可以组成凸四边形

     1 #include <map>
 2 #include <set>
 3 #include <stack>
 4 #include <queue>
 5 #include <cmath>
 6 #include <ctime>
 7 #include <vector>
 8 #include <cstdio>
 9 #include <cctype>
10 #include <cstring>
11 #include <cstdlib>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 #define eps 1e-12
16 #define MAXN 55
17 #define INF 1e30
18 #define mem0(a) memset(a,0, sizeof(a))
19 #define mem1(a) memset(a,-1,sizeof(a))
20 double MAX(double a, double b) {return a > b ? a : b;}
21 double MIn(double a, double b) {return a < b ? a : b;}
22 typedef __int64 LL;
23 /****************************************计算几何头文件**************************************************/
24 struct Point{
25     LL x,y;
26     Point(LL x=0, LL y=0):x(x),y(y){}
27 };
28 
29 Point operator + (Point A, Point B) {return Point(A.x+B.x, A.y+B.y);}
30 
31 Point operator - (Point A, Point B) {return Point(A.x-B.x, A.y-B.y);}
32 
33 LL cross(Point A, Point B) {return A.x*B.y - A.y*B.x;}
34 
35 bool crossed(Point a, Point b, Point c, Point d)//线段ab和cd是否相交
36 {
37     if(cross(a-c, d-c)*cross(b-c, d-c)<0
38        && cross(c-a, b-a)*cross(d-a, b-a)<0)
39     {
40         return true;
41     }
42     return false;
43 }
44 
45 /****************************************************************************************************/
46 
47 int T, N;
48 Point p[100];
49 
50 int is(int a, int b, int c, int d)
51 {
52     if(crossed(p[a], p[b], p[c], p[d])
53        ||crossed(p[a], p[c], p[b], p[d])
54        ||crossed(p[a], p[d], p[b], p[c])) {
55            return 1;
56     }
57     return 0;
58 }
59 
60 int main()
61 {
62     cin >> T;
63     for(int t = 1; t <= T; t ++) {
64         cin >> N;
65         for(int i = 0; i < N; i ++) {
66             scanf("%I64d %I64d", &p[i].x, &p[i].y);
67         }
68         LL ans = 0;
69         for(int i = 0; i < N; i ++) {
70             for(int j = i + 1; j < N; j ++) {
71                 for(int k = j + 1; k < N; k ++) {
72                     for(int l = k + 1; l < N; l ++) {
73                         if(is(i, j, k, l)) ans ++;
74                     }
75                 }
76             }
77         }
78         cout <<"Case " << t << ": ";
79         cout << ans << endl;
80     }
81     return 0;
82 }
    View Code

    F:不会

    G:暴力枚举

      1 #include <map>
  2 #include <set>
  3 #include <stack>
  4 #include <queue>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <vector>
  8 #include <cstdio>
  9 #include <cctype>
 10 #include <cstring>
 11 #include <cstdlib>
 12 #include <iostream>
 13 #include <algorithm>
 14 using namespace std;
 15 #define INF 0x3f3f3f3f
 16 #define inf (-((LL)1<<40))
 17 #define lson k<<1, L, mid
 18 #define rson k<<1|1, mid+1, R
 19 #define mem0(a) memset(a,0,sizeof(a))
 20 #define mem1(a) memset(a,-1,sizeof(a))
 21 #define mem(a, b) memset(a, b, sizeof(a))
 22 #define FOPENIN(IN) freopen(IN, "r", stdin)
 23 #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
 24 #define rep(i, a, b) for(int i = a; i <= b; i ++)
 25 
 26 template<class T> T CMP_MIN(T a, T b) { return a < b; }
 27 template<class T> T CMP_MAX(T a, T b) { return a > b; }
 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
 32 
 33 //typedef __int64 LL;
 34 typedef long long LL;
 35 const int MAXN = 1100;
 36 const int MAXM = 2000010;
 37 const double eps = 1e-10;
 38 
 39 const int dx[4] = {-1, 0, 1, 0};
 40 const int dy[4] = {0, -1, 0, 1};
 41 
 42 int T, M, N;
 43 int vis[20][20];
 44 char ma[20][20];
 45 
 46 struct Point {
 47     int x, y;
 48     Point(){}
 49     Point(int _x, int _y){
 50         x = _x; y = _y;
 51     }
 52 }p[110];
 53 int tot;
 54 
 55 void dfs(int x, int y) {
 56     if(x < 0 || x >= M || y < 0 || y >= N || vis[x][y] || ma[x][y]=='.') return ;
 57     vis[x][y] = 1;
 58     for(int i = 0; i < 4; i ++) {
 59         dfs(x + dx[i], y + dy[i]);
 60     }
 61 }
 62 
 63 int bfs(Point a, Point b)
 64 {
 65     mem1(vis);
 66     queue<Point>q;
 67     q.push(a); q.push(b);
 68     vis[a.x][a.y] = vis[b.x][b.y] = 0;
 69     int ret = 0, cnt = 2;
 70     while(!q.empty()) {
 71         Point fr = q.front(); q.pop();
 72         ret = vis[fr.x][fr.y];
 73         rep (d, 0, 3) {
 74             int x = fr.x + dx[d], y = fr.y + dy[d];
 75             if(x >= 0 && x < M && y >= 0 && y < N && vis[x][y] == -1 && ma[x][y]=='#') {
 76                 q.push(Point(x, y));
 77                 vis[x][y] = vis[fr.x][fr.y] + 1;
 78                 cnt ++;
 79             }
 80         }
 81     }
 82     if(cnt != tot) return INF;
 83     return ret;
 84 }
 85 
 86 int main()
 87 {
 88     cin>>T;
 89     for(int t = 1; t <= T; t ++) {
 90         tot = 0;
 91         printf("Case %d: ", t);
 92         scanf("%d %d%*c", &M, &N);
 93         rep (i, 0, M - 1) {
 94             scanf("%s", ma[i]);
 95             rep (j, 0, N - 1) {
 96                 if(ma[i][j] == '#') p[tot].x = i, p[tot++].y = j;
 97             }
 98         }
 99         if(tot <= 2) {
100             cout<<0<<endl;
101             continue;
102         }
103 
104         mem0(vis);
105         int num = 0;
106         rep (i, 0, M - 1) rep (j, 0, N - 1) if(!vis[i][j] && ma[i][j] == '#') {
107             dfs(i, j);
108             num ++;
109         }
110         if(num == 0 || num > 2) {
111             cout<<-1<<endl;
112             continue;
113         }
114 
115         int ans = INF;
116         rep (i, 0, tot - 1) rep (j, i + 1, tot - 1) {
117             ans = min(ans, bfs(p[i], p[j]));
118         }
119         printf("%d
", ans == INF ? -1 : ans);
120     }
121     return 0;
122 }
    View Code
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  • 原文地址：https://www.cnblogs.com/gj-Acit/p/4374970.html
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