1.hdu 1009 FatMouse' Trade
题目传送门[戳我]
题解传送门[戳我]
2.删数问题
题目&评测传送门[戳我]
（PS）
题解:
通过贪心的策略来想，应该是每一次选择局部最优解。
这个局部最优解就是使当前的数最小，所以可以从高位到低位进行查找，使得s[j + 1] >= s[j]否则，删掉第j个数
Code
1 /** 2 * smartoj 3 * Problem#1098 4 * Accepted 5 * Time:107ms 6 * Memory:516k 7 */ 8 #include<iostream> 9 #include<cstdio> 10 #include<cctype> 11 #include<cstring> 12 #include<cstdlib> 13 #include<cmath> 14 #include<fstream> 15 #include<sstream> 16 #include<algorithm> 17 #include<map> 18 #include<set> 19 #include<queue> 20 #include<vector> 21 #include<stack> 22 using namespace std; 23 typedef bool boolean; 24 #define INF 0xfffffff 25 #define smin(a, b) a = min(a, b) 26 #define smax(a, b) a = max(a, b) 27 template<typename T> 28 inline void readInteger(T& u){ 29 char x; 30 int aFlag = 1; 31 while(!isdigit((x = getchar())) && x != '-'); 32 if(x == '-'){ 33 x = getchar(); 34 aFlag = -1; 35 } 36 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 37 ungetc(x, stdin); 38 u *= aFlag; 39 } 40 41 vector<char> str; 42 int s; 43 44 inline void init(){ 45 char x; 46 while((x = getchar()) != ' '){ 47 str.push_back(x); 48 } 49 readInteger(s); 50 } 51 52 inline void solve(){ 53 for(int i = 1; i <= s; i++){ 54 int j = 0; 55 while(j < (signed)str.size() - 1 && str[j + 1] >= str[j]){ 56 j++; 57 } 58 str.erase(str.begin() + j); 59 } 60 int i = 0; 61 while(i < str.size() && str[i] == '0') i++; 62 for(; i < str.size(); i++){ 63 putchar(str[i]); 64 } 65 } 66 67 int main(){ 68 init(); 69 solve(); 70 return 0; 71 }
3.hdu 1045 Fire Net
题目传送门[戳我]
题解:
一个很神奇的方法AC，按如果在(x, y)这个地方建城堡造成的影响数排序，然后for循环一边，把能建的都建，然后输出结果。
Code
1 /** 2 * acm.hdu.com.cn 3 * Problem#1045 4 * Accepted 5 * Time:15ms 6 * Memory:1568k 7 */ 8 #include<iostream> 9 #include<cstdio> 10 #include<cctype> 11 #include<cstring> 12 #include<cstdlib> 13 #include<cmath> 14 #include<fstream> 15 #include<sstream> 16 #include<algorithm> 17 #include<map> 18 #include<set> 19 #include<queue> 20 #include<vector> 21 #include<stack> 22 using namespace std; 23 typedef bool boolean; 24 #define INF 0xfffffff 25 #define smin(a, b) a = min(a, b) 26 #define smax(a, b) a = max(a, b) 27 template<typename T> 28 inline void readInteger(T& u){ 29 char x; 30 int aFlag = 1; 31 while(!isdigit((x = getchar())) && x != '-'); 32 if(x == '-'){ 33 x = getchar(); 34 aFlag = -1; 35 } 36 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 37 ungetc(x, stdin); 38 u *= aFlag; 39 } 40 41 int n; 42 int f[5][5]; 43 44 inline boolean init(){ 45 readInteger(n); 46 if(n == 0) return false; 47 for(int i = 1; i <= n; i++){ 48 getchar(); 49 for(int j = 1; j <= n; j++){ 50 char x = getchar(); 51 if(x == 'X') 52 f[i][j] = 1; 53 else f[i][j] = 0; 54 } 55 } 56 return true; 57 } 58 59 int roles[5][5]; 60 61 void fire(int x, int y){ 62 for(int i = x - 1; i >= 1 && f[i][y] == 0; i--){ roles[x][y]++; } 63 for(int i = x + 1; i <= n && f[i][y] == 0; i++){ roles[x][y]++; } 64 for(int j = y - 1; j >= 1 && f[x][j] == 0; j--){ roles[x][y]++; } 65 for(int j = y + 1; j <= n && f[x][j] == 0; j++){ roles[x][y]++; } 66 } 67 68 typedef class Data{ 69 public: 70 int x, y; 71 int data; 72 Data(const int x = 0, const int y = 0, const int data = 0):x(x), y(y), data(data){} 73 }Data; 74 75 boolean cmpare(const Data& a, const Data& b){ 76 return a.data < b.data; 77 } 78 79 vector<Data> l; 80 boolean visited[5][5]; 81 inline void solve(){ 82 memset(roles, 1, sizeof(roles)); 83 memset(visited, false, sizeof(visited)); 84 for(int i = 1; i <= n; i++){ 85 for(int j = 1; j <= n; j++){ 86 if(f[i][j] == 0){ 87 fire(i, j); 88 l.push_back(Data(i, j, roles[i][j])); 89 } 90 } 91 } 92 sort(l.begin(), l.end(), cmpare); 93 int result = 0; 94 for(int k = 0; k < (signed)l.size(); k++){ 95 if(!visited[l[k].x][l[k].y]){ 96 result++; 97 int &x = l[k].x, &y= l[k].y; 98 for(int i = x - 1; i >= 1 && f[i][y] == 0; i--){ visited[i][y] = true; } 99 for(int i = x + 1; i <= n && f[i][y] == 0; i++){ visited[i][y] = true; } 100 for(int j = y - 1; j >= 1 && f[x][j] == 0; j--){ visited[x][j] = true; } 101 for(int j = y + 1; j <= n && f[x][j] == 0; j++){ visited[x][j] = true; } 102 } 103 } 104 l.clear(); 105 printf("%d ", result); 106 } 107 108 int main(){ 109 while(init()){ 110 solve(); 111 } 112 return 0; 113 }
4.hdu 1050 Moving Table
题目传送门[戳我]
题解:
还是很神奇的方法，用差分，标记每次移动影响的范围，前缀和完了找出一个最大值乘10。因为不同地方可以同时进行，只有有地方重叠的时候需要分批扮桌子。
Code
1 /** 2 * acm.hdu.com.cn 3 * Problem#1050 4 * Accepted 5 * Time:0ms 6 * Memory:1552k 7 */ 8 #include<iostream> 9 #include<cstdio> 10 #include<cctype> 11 #include<cstring> 12 #include<cstdlib> 13 #include<fstream> 14 #include<sstream> 15 #include<algorithm> 16 #include<map> 17 #include<set> 18 #include<queue> 19 #include<vector> 20 #include<stack> 21 #include<cmath> 22 using namespace std; 23 typedef bool boolean; 24 #define INF 0xfffffff 25 #define smin(a, b) a = min(a, b) 26 #define smax(a, b) a = max(a, b) 27 template<typename T> 28 inline void readInteger(T& u){ 29 char x; 30 int aFlag = 1; 31 while(!isdigit((x = getchar())) && x != '-'); 32 if(x == '-'){ 33 x = getchar(); 34 aFlag = -1; 35 } 36 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 37 ungetc(x, stdin); 38 u *= aFlag; 39 } 40 41 int T; 42 int n; 43 int used[205]; 44 45 inline void init(){ 46 readInteger(n); 47 for(int i = 1, a, b, c, d; i <= n; i++){ 48 readInteger(a); 49 readInteger(b); 50 c = (min(a, b) + 1) >> 1; 51 d = (max(a, b) + 1) >> 1; 52 used[c] += 1, used[d + 1] -= 1; 53 } 54 } 55 56 inline void solve(){ 57 int result = 0; 58 for(int i = 1; i <= 201; i++){ 59 used[i] += used[i - 1]; 60 smax(result, used[i]); 61 } 62 printf("%d ", result * 10); 63 memset(used, 0, sizeof(used)); 64 } 65 66 int main(){ 67 readInteger(T); 68 while(T--){ 69 init(); 70 solve(); 71 } 72 return 0; 73 }
5.hdu 1051 Wooden Sticks
题目传送门[here]
题解:
当成一个二元组排个序，然后每次往后找，保证每次找到的都不需要准备时间，每次循环完了将准备时间加1，前面最大的weight清0
Code
1 /** 2 * acm.hdu.com.cn 3 * Problem#1051 4 * Accepted 5 * Time:15ms 6 * Memory:1644k 7 */ 8 #include<iostream> 9 #include<cstdio> 10 #include<cctype> 11 #include<cstring> 12 #include<cstdlib> 13 #include<cmath> 14 #include<fstream> 15 #include<sstream> 16 #include<algorithm> 17 #include<map> 18 #include<set> 19 #include<queue> 20 #include<vector> 21 #include<stack> 22 using namespace std; 23 typedef bool boolean; 24 #define INF 0xfffffff 25 #define smin(a, b) a = min(a, b) 26 #define smax(a, b) a = max(a, b) 27 template<typename T> 28 inline void readInteger(T& u){ 29 char x; 30 int aFlag = 1; 31 while(!isdigit((x = getchar())) && x != '-'); 32 if(x == '-'){ 33 x = getchar(); 34 aFlag = -1; 35 } 36 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 37 ungetc(x, stdin); 38 u *= aFlag; 39 } 40 41 typedef class Stick{ 42 public: 43 int length; 44 int weight; 45 Stick(const int length = 0, const int weight = 0):length(length), weight(weight){ } 46 boolean operator <(Stick another) const { 47 if(this->length != another.length) return this->length < another.length; 48 return this->weight < another.weight; 49 } 50 }Stick; 51 52 int T; 53 int n; 54 Stick *ss; 55 56 inline void init(){ 57 readInteger(n); 58 ss = new Stick[(const int)(n + 1)]; 59 for(int i = 1; i <= n; i++){ 60 readInteger(ss[i].length); 61 readInteger(ss[i].weight); 62 } 63 } 64 65 boolean *visited; 66 inline void solve(){ 67 int finished = 0; 68 int result = 0, last; 69 sort(ss + 1, ss + n + 1); 70 visited = new boolean[(const int)(n + 1)]; 71 memset(visited, false, sizeof(boolean) * (n + 1)); 72 while(finished < n){ 73 result++; 74 last = 0; 75 for(int i = 1; i <= n; i++){ 76 if(!visited[i] && ss[i].weight >= last){ 77 finished++; 78 visited[i] = true; 79 last = ss[i].weight; 80 } 81 } 82 } 83 printf("%d", result); 84 delete[] ss; 85 delete[] visited; 86 } 87 88 int main(){ 89 readInteger(T); 90 while(T--){ 91 init(); 92 solve(); 93 putchar(' '); 94 } 95 return 0; 96 }
6.UVa 10755 Garbage Heap
原题传送门[here]
VJudge传送门:[here]
题解:
首先降低维数，枚举x1, x2, y1, y2，然后可以将这一段压缩成一维的一段。然后就可以贪心了，为了让z1 - z2这一段竟可能大，所以应该使s[z2] - s[z1 - 1]尽可能小。然后加个前缀和降低时间复杂度，就可以开开心心地交vjudge了
1 /** 2 * uva 3 * Problem#10755 4 * Accepted 5 * Time:180ms 6 */ 7 #include<iostream> 8 #include<cstdio> 9 #include<cctype> 10 #include<cstring> 11 #include<cstdlib> 12 #include<cmath> 13 #include<fstream> 14 #include<sstream> 15 #include<algorithm> 16 #include<map> 17 #include<set> 18 #include<queue> 19 #include<vector> 20 #include<stack> 21 using namespace std; 22 typedef bool boolean; 23 #define INF 0xfffffff 24 #define smin(a, b) a = min(a, b) 25 #define smax(a, b) a = max(a, b) 26 template<typename T> 27 inline void readInteger(T& u){ 28 char x; 29 int aFlag = 1; 30 while(!isdigit((x = getchar())) && x != '-'); 31 if(x == '-'){ 32 x = getchar(); 33 aFlag = -1; 34 } 35 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 36 ungetc(x, stdin); 37 u *= aFlag; 38 } 39 40 const long long inf = (1LL << 60); 41 42 template<typename T> 43 inline void putInteger(T u){ 44 if(u == 0){ 45 putchar('0'); 46 return; 47 } 48 if(u < 0){ 49 putchar('-'); 50 u *= -1; 51 } 52 stack<char> s; 53 while(u != 0) s.push(u % 10 + '0'), u /= 10; 54 while(!s.empty()) putchar(s.top()), s.pop(); 55 } 56 57 void expand(int i, int& x, int& y, int& z, int& sign){ //容斥原理 58 x = i & 1, i >>= 1; 59 y = i & 1, i >>= 1; 60 z = i & 1; 61 sign = ((x + y + z) % 2 == 1) ? (1) : (-1); 62 } 63 64 int T; 65 int a, b, c; 66 67 long long s[21][21][21]; 68 inline void init_sum(){ 69 for(int i = 1; i <= a; i++){ 70 for(int j = 1; j <= b; j++){ 71 for(int k = 1; k <= c; k++){ 72 for(int p = 1; p < 8; p++){ 73 int x, y, z, sign; 74 expand(p, x, y, z, sign); 75 s[i][j][k] += s[i - x][j - y][k - z] * sign; 76 } 77 } 78 } 79 } 80 } 81 82 //long long f[21]; 83 inline long long dp(int x1, int x2, int y1, int y2){ 84 long long sumz; 85 long long result = -inf; 86 long long minv = 0; 87 int dx = x2 - x1 + 1, dy = y2 - y1 + 1; 88 for(int i = 1; i <= c; i++){ 89 sumz = s[x2][y2][i]; 90 for(int p = 1; p < 8; p++){ 91 int x, y, z, sign; 92 expand(p, x, y, z, sign); 93 //sumz -= s[x2 - x * dx][y2 - y * dy][i - z] * sign; 94 sumz -= s[x2 - x * dx][y2 - y * dy][i - z * i] * sign; 95 } 96 smax(result , sumz - minv); 97 smin(minv, sumz); 98 // f[i] = max(0LL, f[i - 1]) + sumz; 99 // smax(result, f[i]); 100 } 101 return result; 102 } 103 104 inline void init(){ 105 readInteger(a); 106 readInteger(b); 107 readInteger(c); 108 memset(s, 0, sizeof(s)); 109 for(int i = 1; i <= a; i++){ 110 for(int j = 1; j <= b; j++){ 111 for(int k = 1; k <= c; k++){ 112 readInteger(s[i][j][k]); 113 } 114 } 115 } 116 } 117 118 inline void solve(){ 119 long long result = -inf; 120 for(int x1 = 1; x1 <= a; x1++){ 121 for(int x2 = x1; x2 <= a; x2++){ 122 for(int y1 = 1; y1 <= b; y1++){ 123 for(int y2 = y1; y2 <= b; y2++){ 124 long long cmp = dp(x1, x2, y1, y2); 125 smax(result, cmp); 126 } 127 } 128 } 129 } 130 putInteger(result); 131 putchar(' '); 132 } 133 134 int main(){ 135 readInteger(T); 136 while(T--){ 137 init(); 138 init_sum(); 139 solve(); 140 if(T) putchar(' '); 141 } 142 return 0; 143 }
7.USACO 1.3 Mixing Milk
原题传送门[here]
题目大意:
一个公司需要购买N个单位的牛奶，有M个农民每天提供定价定量的牛奶。第i个农民的牛奶每个单位售价为Pi，每天提供Ai个单位的牛奶。求每天最小的开销。
题解:
和第一道题有些相似，按照单价排序，然后计算结果就行了。
1 /* 2 ID: m1582851 3 PROG: milk 4 LANG: C++11 5 */ 6 /** 7 * USACO 8 * Accepted 9 * Time:0ms 10 * Memory:4184k 11 */ 12 #include<iostream> 13 #include<cstdio> 14 #include<cctype> 15 #include<cstring> 16 #include<cstdlib> 17 #include<fstream> 18 #include<sstream> 19 #include<algorithm> 20 #include<map> 21 #include<set> 22 #include<queue> 23 #include<vector> 24 #include<stack> 25 using namespace std; 26 typedef bool boolean; 27 #define INF 0xfffffff 28 #define smin(a, b) a = min(a, b) 29 #define smax(a, b) a = max(a, b) 30 template<typename T> 31 inline void readInteger(T& u){ 32 char x; 33 int aFlag = 1; 34 while(!isdigit((x = getchar())) && x != '-'); 35 if(x == '-'){ 36 x = getchar(); 37 aFlag = -1; 38 } 39 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 40 ungetc(x, stdin); 41 u *= aFlag; 42 } 43 44 typedef class Milk{ 45 public: 46 int weight; 47 int cost; 48 Milk(const int weight = 0, const int cost = 0):weight(weight), cost(cost){ } 49 boolean operator < (Milk another) const { 50 return this->cost < another.cost; 51 } 52 }Milk; 53 54 int n, m; 55 Milk* milks; 56 int result = 0; 57 58 inline void init(){ 59 readInteger(n); 60 readInteger(m); 61 milks = new Milk[(const int)(m + 1)]; 62 for(int i = 1; i <= m; i++){ 63 readInteger(milks[i].cost); 64 readInteger(milks[i].weight); 65 } 66 } 67 68 inline void solve(){ 69 sort(milks + 1, milks + m + 1); 70 for(int i = 1; i <= m && n; i++){ 71 int maxcer = min(n, milks[i].weight); 72 result += maxcer * milks[i].cost; 73 n -= maxcer; 74 } 75 printf("%d ", result); 76 } 77 78 int main(){ 79 freopen("milk.in", "r", stdin); 80 freopen("milk.out", "w", stdout); 81 init(); 82 solve(); 83 return 0; 84 }
8.USACO 1.3 Barn Repair
原题传送门[here]
题目大意:
用长度任意，数量不超过M的木板挡住有牛的牛棚（每个牛棚的长度都看作1， 一共有C个牛棚有牛，并且给出每只牛在哪个牛棚里）的出路（必须堵住所有牛棚的出口），求最小的木板的长度的和
题解:
首先把一段一段的连续有牛的牛棚之间的间隔的长度求出来，计算的方法是后一个的起始位置减去前一个的结束位置减1。只需要连接间隔较小的几段，所以就从小到大排序。需要连接的次数是 总空缺数减去M然后加1。接着把最小的几个加起来然后加上C就完成了。
1 /* 2 ID: m1582851 3 PROG: barn1 4 LANG: C++11 5 */ 6 /** 7 * USACO 8 * Accepted 9 * Time:0ms 10 * Memory:4184k 11 */ 12 #include<iostream> 13 #include<cstdio> 14 #include<cctype> 15 #include<cstring> 16 #include<cstdlib> 17 #include<fstream> 18 #include<sstream> 19 #include<algorithm> 20 #include<map> 21 #include<set> 22 #include<queue> 23 #include<vector> 24 #include<stack> 25 using namespace std; 26 typedef bool boolean; 27 #define INF 0xfffffff 28 #define smin(a, b) a = min(a, b) 29 #define smax(a, b) a = max(a, b) 30 template<typename T> 31 inline void readInteger(T& u){ 32 char x; 33 int aFlag = 1; 34 while(!isdigit((x = getchar())) && x != '-'); 35 if(x == '-'){ 36 x = getchar(); 37 aFlag = -1; 38 } 39 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 40 ungetc(x, stdin); 41 u *= aFlag; 42 } 43 44 int m, s, c; 45 vector<int> jgs; 46 int* stalls; 47 int result; 48 49 inline void init(){ 50 readInteger(m); 51 readInteger(s); 52 readInteger(c); 53 stalls = new int[(const int)(c + 1)]; 54 for(int i = 1; i <= c; i++){ 55 readInteger(stalls[i]); 56 } 57 } 58 59 inline void solve(){ 60 sort(stalls + 1, stalls + c + 1); 61 for(int i = 2; i <= c; i++){ 62 if(stalls[i] > stalls[i - 1] + 1) 63 jgs.push_back(stalls[i] - stalls[i - 1] - 1); 64 } 65 sort(jgs.begin(), jgs.end()); 66 int needed = (signed)jgs.size() - m; 67 result = c; 68 for(int i = 0; i <= needed; i++){ 69 result += jgs[i]; 70 } 71 printf("%d ", result); 72 } 73 74 int main(){ 75 freopen("barn1.in", "r", stdin); 76 freopen("barn1.out", "w", stdout); 77 init(); 78 solve(); 79 return 0; 80 }