第一题比较简单，用exist数组判断是否在循环队列中，就可实现线性算法。
Code
1 #include<iostream>
2 #include<cstdio>
3 #include<cctype>
4 #include<cstring>
5 #include<cstdlib>
6 #include<fstream>
7 #include<sstream>
8 #include<algorithm>
9 #include<map>
10 #include<set>
11 #include<queue>
12 #include<vector>
13 #include<stack>
14 using namespace std;
15 typedef bool boolean;
16 #define INF 0xfffffff
17 #define smin(a, b) a = min(a, b)
18 #define smax(a, b) a = max(a, b)
19 template<typename T>
20 inline void readInteger(T& u){
21 char x;
22 int aFlag = 1;
23 while(!isdigit((x = getchar())) && x != ‘-‘);
24 if(x == ‘-‘){
25 x = getchar();
26 aFlag = -1;
27 }
28 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘);
29 ungetc(x, stdin);
30 u *= aFlag;
31 }
32
33 int n, m;
34 int top;
35 int *s;
36 boolean exist[1002];
37
38 inline void s_push(int* sta, int x) {
39 exist[sta[top] + 1] = false;
40 sta[top] = x;
41 exist[x + 1] = true;
42 if(top == m - 1) top = 0;
43 else top = top + 1;
44 }
45
46 int res = 0;
47 inline void solve() {
48 readInteger(m);
49 readInteger(n);
50 s = new int[(const int)(m + 1)];
51 memset(s, -1, sizeof(int) * (m + 1));
52 top = 0;
53 for(int i = 1, a; i <= n; i++) {
54 readInteger(a);
55 if(!exist[a + 1]) {
56 res++;
57 s_push(s, a);
58 }
59 }
60 printf("%d", res);
61 }
62
63 int main() {
64 freopen("translate.in", "r", stdin);
65 freopen("translate.out", "w", stdout);
66 solve();
67 return 0;
68 }
因为当卡牌的选择数是固定的情况下，跳的长度也就知道了，于是就可以用四种卡牌来做状态，得到了dp方程:
为了防止过多的无用的状态，所以就用记忆化搜索(其实直接4个for也没什么问题)
Code
1 #include<iostream>
2 #include<cstdio>
3 #include<cctype>
4 #include<cstring>
5 #include<cstdlib>
6 #include<fstream>
7 #include<sstream>
8 #include<algorithm>
9 #include<map>
10 #include<set>
11 #include<queue>
12 #include<vector>
13 #include<stack>
14 using namespace std;
15 typedef bool boolean;
16 #define INF 0xfffffff
17 #define smin(a, b) a = min(a, b)
18 #define smax(a, b) a = max(a, b)
19 template<typename T>
20 inline void readInteger(T& u){
21 char x;
22 int aFlag = 1;
23 while(!isdigit((x = getchar())) && x != ‘-‘);
24 if(x == ‘-‘){
25 x = getchar();
26 aFlag = -1;
27 }
28 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘);
29 ungetc(x, stdin);
30 u *= aFlag;
31 }
32
33 int n, m;
34 int counter[4];
35 int f[41][41][41][41];
36 boolean vis[41][41][41][41];
37 int *val;
38
39 inline void init() {
40 readInteger(n);
41 readInteger(m);
42 memset(counter, 0, sizeof(counter));
43 memset(f, 0, sizeof(f));
44 memset(vis, false, sizeof(vis));
45 val = new int[(const int)(n + 1)];
46 for(int i = 1; i <= n; i++) readInteger(val[i]);
47 for(int i = 1, a; i <= m; i++) {
48 readInteger(a);
49 counter[a - 1]++;
50 }
51 f[0][0][0][0] = val[1];
52 vis[0][0][0][0] = true;
53 }
54
55 int dfs(int a, int b, int c, int d) {
56 if(vis[a][b][c][d]) return f[a][b][c][d];
57 vis[a][b][c][d] = true;
58 int pos = a * 1 + b * 2 + c * 3 + d * 4 + 1;
59 int ra = 0, rb = 0, rc = 0, rd = 0;
60 if(a > 0) ra = dfs(a - 1, b, c, d);
61 if(b > 0) rb = dfs(a, b - 1, c, d);
62 if(c > 0) rc = dfs(a, b, c - 1, d);
63 if(d > 0) rd = dfs(a, b, c, d - 1);
64 smax(f[a][b][c][d], max(ra, max(rb, max(rc, rd))) + val[pos]);
65 return f[a][b][c][d];
66 }
67
68 inline void solve() {
69 int res = dfs(counter[0], counter[1], counter[2], counter[3]);
70 printf("%d", res);
71 }
72
73 int main() {
74 freopen("tortoise.in", "r", stdin);
75 freopen("tortoise.out", "w", stdout);
76 init();
77 solve();
78 return 0;
79 }
这道题相当于是安排罪犯，使所有的影响力的最大值最小，于是就不难想到二分答案。
对于当前二分值mid，无视所有权值小于等于它的边，判断是否能让剩下的边构成二分图。至于这一步多次对未访问的点进行dfs，将与它相连的点丢进另一个监狱，如果出现矛盾，就说明当前的二分值不合法。
Code
1 #include<iostream>
2 #include<cstdio>
3 #include<cctype>
4 #include<cstring>
5 #include<cstdlib>
6 #include<fstream>
7 #include<sstream>
8 #include<algorithm>
9 #include<map>
10 #include<set>
11 #include<queue>
12 #include<vector>
13 #include<stack>
14 using namespace std;
15 typedef bool boolean;
16 #define INF 0xfffffff
17 #define smin(a, b) a = min(a, b)
18 #define smax(a, b) a = max(a, b)
19 template<typename T>
20 inline void readInteger(T& u){
21 char x;
22 int aFlag = 1;
23 while(!isdigit((x = getchar())) && x != ‘-‘);
24 if(x == ‘-‘){
25 x = getchar();
26 aFlag = -1;
27 }
28 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘);
29 ungetc(x, stdin);
30 u *= aFlag;
31 }
32
33 typedef class Edge {
34 public:
35 int end;
36 int next;
37 int w;
38 Edge(const int end = 0, const int next = 0, const int w = 0):end(end), next(next), w(w) { }
39 }Edge;
40
41 typedef class MapManager{
42 public:
43 int ce;
44 int *h;
45 Edge *edge;
46 MapManager(){}
47 MapManager(int points, int limit):ce(0){
48 h = new int[(const int)(points + 1)];
49 edge = new Edge[(const int)(limit + 1)];
50 memset(h, 0, sizeof(int) * (points + 1));
51 }
52 inline void addEdge(int from, int end, int w){
53 edge[++ce] = Edge(end, h[from], w);
54 h[from] = ce;
55 }
56 inline void addDoubleEdge(int from, int end, int w){
57 addEdge(from, end, w);
58 addEdge(end, from, w);
59 }
60 Edge& operator [](int pos) {
61 return edge[pos];
62 }
63 }MapManager;
64 #define m_begin(g, i) (g).h[(i)]
65
66 int n, m;
67 MapManager g;
68 int* belong;
69 int maxval = -1;
70
71 inline void init() {
72 readInteger(n);
73 readInteger(m);
74 g = MapManager(n, 2 * m);
75 belong = new int[(const int)(n + 1)];
76 for(int i = 1, a, b, w; i <= m; i++) {
77 readInteger(a);
78 readInteger(b);
79 readInteger(w);
80 g.addDoubleEdge(a, b, w);
81 smax(maxval, w);
82 }
83 }
84
85 boolean dfs(int node, int val, int x) {
86 if(belong[node] != -1) return belong[node] == val;
87 else belong[node] = val;
88 for(int i = m_begin(g, node); i != 0; i = g[i].next) {
89 if(g[i].w <= x) continue;
90 int& e = g[i].end;
91 if(!dfs(e, val ^ 1, x)) return false;
92 }
93 return true;
94 }
95
96 boolean check(int x) {
97 memset(belong, -1, sizeof(int) * (n + 1));
98 for(int i = 1; i <= n; i++) {
99 if(belong[i] == -1)
100 if(!dfs(i, 0, x)) return false;
101 }
102 return true;
103 }
104
105 inline void solve() {
106 int l = 0, r = maxval;
107 while(l <= r) {
108 int mid = (l + r) >> 1;
109 if(!check(mid)) l = mid + 1;
110 else r = mid - 1;
111 }
112 printf("%d", r + 1);
113 }
114
115 int main() {
116 freopen("prison.in", "r", stdin);
117 freopen("prison.out", "w", stdout);
118 init();
119 solve();
120 return 0;
121 }
对于可以使干旱区的每个城市都能够获得水的情况，那么对于湖泊边的某个城市建的蓄水厂，能够提供给干旱区的城市水的集合，是一段连续的区间。这里可以简要地说明一下
假设上面的命题不成立，就会出现形如上面的情况，那么为了使中间的小圆圈所在的城市可以得到供水，那么一定会有其他的水路和这几条水路交叉，也就是说这个蓄水厂还是可以给小圆圈所在的城市供水，矛盾，所以上面的命题成立。
于是我们可以通过M次dfs得到假设在第一行第i个城市建蓄水厂所能够供水的区间，当然这样的时间复杂度最极端的情况下是O(m2n)。但是仔细一想,对于一个点(i,j)它可以到达的干旱区城市，那么某个可以到达它的点(i+a,j+b)一定也可以到达它所可以到达的干旱区城市，于是只需要一些dfs从没有被访问的第一行的城市开始搜索，因为每个城市只会被访问一次，所以时间复杂度为O(mn)，降回了可以接受的范围。
得到了这么一堆区间可以干什么?我们是需要选择一些区间使得它们的并集是整个干旱区，于是可以用dp[i]来表示完全覆盖干旱区第1个城市到第j个城市的最小所需的区间，于是又得到了方程:
Code
1 #include<iostream>
2 #include<cstdio>
3 #include<cctype>
4 #include<cstring>
5 #include<cstdlib>
6 #include<fstream>
7 #include<sstream>
8 #include<algorithm>
9 #include<map>
10 #include<set>
11 #include<queue>
12 #include<vector>
13 #include<stack>
14 using namespace std;
15 typedef bool boolean;
16 #define INF 0xfffffff
17 #define smin(a, b) a = min(a, b)
18 #define smax(a, b) a = max(a, b)
19 template<typename T>
20 inline void readInteger(T& u){
21 char x;
22 int aFlag = 1;
23 while(!isdigit((x = getchar())) && x != ‘-‘);
24 if(x == ‘-‘){
25 x = getchar();
26 aFlag = -1;
27 }
28 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘);
29 ungetc(x, stdin);
30 u *= aFlag;
31 }
32
33 typedef class Point {
34 public:
35 int x, y;
36 int l, r;
37
38 Point(const int x = 0, const int y = 0, const int l = 0, const int r = 0):x(x), y(y), l(l), r(r) { }
39
40 boolean mergable(Point& b) {
41 if(b.l < l) return true;
42 if(b.r > r) return true;
43 return false;
44 }
45 void merge(Point& b) {
46 smin(l, b.l);
47 smax(r, b.r);
48 }
49 }Point;
50
51 int n, m;
52 int mat[505][505];
53
54 inline void init() {
55 readInteger(n);
56 readInteger(m);
57 for(int i = 1; i <= n; i++) {
58 for(int j = 1; j <= m; j++) {
59 readInteger(mat[i][j]);
60 }
61 }
62 }
63
64 const int mover[2][4] = {{0, 1, 0, -1}, {1, 0, -1, 0}};
65
66 Point f[505][505];
67 boolean vis[505][505];
68 Point dfs(int x, int y) {
69 if(vis[x][y]) return f[x][y];
70 vis[x][y] = true;
71 f[x][y] = Point(x, y, 999, 0);
72 if(x == n) f[x][y] = Point(x, y, y, y);
73 for(int k = 0; k < 4; k++) {
74 int x1 = x + mover[0][k], y1 = y + mover[1][k];
75 if(x1 < 1 || x1 > n) continue;
76 if(y1 < 1 || y1 > m) continue;
77 if(mat[x1][y1] < mat[x][y]) {
78 Point e = dfs(x1, y1);
79 f[x][y].merge(e);
80 }
81 }
82 return f[x][y];
83 }
84
85 void dfs1(int x, int y) {
86 if(vis[x][y]) return;
87 vis[x][y] = true;
88 for(int k = 0; k < 4; k++) {
89 int x1 = x + mover[0][k], y1 = y + mover[1][k];
90 if(x1 < 1 || x1 > n) continue;
91 if(y1 < 1 || y1 > m) continue;
92 if(mat[x1][y1] < mat[x][y]) {
93 Point e = dfs(x1, y1);
94 f[x][y].merge(e);
95 }
96 }
97 }
98
99 int *dp;
100 inline void solve() {
101 memset(vis, false, sizeof(vis));
102 for(int i = 1; i <= m; i++)
103 if(!vis[1][i])
104 dfs(1, i);
105 dp = new int[(const int)(m + 2)];
106 memset(dp, 0, sizeof(int) * (m + 2));
107 for(int i = 1; i <= m; i++)
108 if(f[1][i].l < 999)
109 dp[f[1][i].l] += 1, dp[f[1][i].r + 1] -= 1;
110 int unget = 0;
111 for(int i = 1; i <= m; i++) {
112 if(!vis[n][i])
113 unget++;
114 }
115 if(unget > 1) {
116 printf("0\n%d", unget);
117 return;
118 }
119 memset(dp, 0x7f, sizeof(int) * (m + 2));
120 dp[0] = 0;
121 for(int i = 1; i <= m; i++) {
122 for(int j = 1; j <= m; j++) {
123 if(f[1][j].l > i || f[1][j].r < i) continue;
124 smin(dp[i], dp[f[1][j].l - 1] + 1);
125 }
126 }
127 printf("1\n%d", dp[m]);
128 }
129
130 int main() {
131 freopen("flow.in", "r", stdin);
132 freopen("flow.out", "w", stdout);
133 init();
134 solve();
135 return 0;
136 }
时间： 2024-10-26 02:31:49