优先队列题目整理
1 /* 2 ***********Good LUCK********** 3 * Author: yeahpeng 4 * Created Time: 2015/6/5 12:32:45 5 * File Name: uestc_482_CExchange.cpp 6 * Charitable Exchange 7 Time Limit: 4000/2000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others) 8 9 Submit Status 10 Have you ever heard a star charity show called Charitable Exchange? In this show, a famous star starts with a small item which values 1 yuan. Then, through the efforts of repeatedly exchanges which continuously increase the value of item in hand, he (she) finally brings back a valuable item and donates it to the needy. 11 12 In each exchange, one can exchange for an item of Vi yuan if he (she) has an item values more than or equal to Ri yuan, with a time cost of Ti minutes. 13 14 Now, you task is help the star to exchange for an item which values more than or equal to M yuan with the minimum time. 15 16 Input 17 The first line of the input is T (no more than 20), which stands for the number of test cases you need to solve. 18 19 For each case, two integers N, M (1¡ÜN¡Ü105, 1¡ÜM¡Ü109) in the first line indicates the number of available exchanges and the expected value of final item. Then N lines follow, each line describes an exchange with 3 integers Vi, Ri, Ti (1¡ÜRi¡ÜVi¡Ü109, 1¡ÜTi¡Ü109). 20 21 Output 22 For every test case, you should output Case #k: first, where k indicates the case number and counts from 1. Then output the minimum time. Output ?1 if no solution can be found. 23 24 Sample input and output 25 Sample Input Sample Output 26 3 27 3 10 28 5 1 3 29 8 2 5 30 10 9 2 31 4 5 32 2 1 1 33 3 2 1 34 4 3 1 35 8 4 1 36 5 9 37 5 1 1 38 10 4 10 39 8 1 10 40 11 6 1 41 7 3 8 42 Case #1: -1 43 Case #2: 4 44 Case #3: 10 45 */ 46 #include <stdio.h> 47 #include <iostream> 48 #include <algorithm> 49 #include <sstream> 50 #include <stdlib.h> 51 #include <string.h> 52 #include <limits.h> 53 #include <vector> 54 #include <string> 55 #include <time.h> 56 #include <math.h> 57 #include <queue> 58 #include <stack> 59 #include <set> 60 #include <map> 61 #define INF 0x3f3f3f3f 62 #define Zero(x) memset((x),0, sizeof(x)) 63 #define Neg(x) memset((x), -1, sizeof(x)) 64 #define dg(x) cout << #x << " = " << x << endl 65 #define pk(x) push_back(x) 66 #define pok() pop_back() 67 #define eps 1e-8 68 #define pii pair<int, int> 69 #define pi acos(-1.0) 70 using namespace std; 71 typedef long long ll; 72 bool debug = true; 73 int OK = 1; 74 const int maxn = 100050; 75 int n, m, c; 76 struct node{ 77 int v, u, val; 78 node(int vv = 0, int uu = 0, int vval = 0){ 79 v = vv; 80 u = uu; 81 val = vval; 82 } 83 bool operator < (const node& a)const { 84 return u < a.u; 85 } 86 }edge[maxn]; 87 ll ans; 88 struct noo{ 89 int u; 90 ll cost; 91 noo(int uu = 0, ll cc = 0){ 92 u = uu; 93 cost = cc; 94 } 95 bool operator < (const noo& a)const { 96 return cost > a.cost; 97 } 98 }; 99 bool cal(){ 100 priority_queue<noo> pq; 101 pq.push(noo(1, 0)); 102 noo top; 103 int left = 0, i; 104 while(!pq.empty()){ 105 top = pq.top(); 106 pq.pop(); 107 if(top.u >= m) { 108 ans = top.cost; 109 return true; 110 } 111 for(i = left; i < c; ++i){ 112 if(top.u < edge[i].u) break; 113 if(top.u >= edge[i].u && top.u <edge[i].v){ 114 pq.push(noo(edge[i].v, top.cost + edge[i].val)); 115 } 116 } 117 left = i; 118 } 119 return false; 120 } 121 122 int main(){ 123 //freopen("data.in","r",stdin); 124 //freopen("data.out","w",stdout); 125 //cin.sync_with_stdio(false); 126 int T; 127 cin >> T; 128 int u, v, val; 129 for(int cs = 1; cs <= T; ++cs){ 130 scanf("%d%d", &n, &m); 131 c = 0; 132 for(int i = 0; i < n; ++i){ 133 scanf("%d%d%d", &v, &u, &val); 134 if(u >= v) continue; 135 edge[c++] = node(v, u, val); 136 } 137 sort(edge, edge + c); 138 printf("Case #%d: ", cs); 139 if(cal()){ 140 cout << ans << endl; 141 }else cout << -1 << endl; 142 } 143 return 0; 144 }
1 /* 2 ***********Good LUCK********** 3 * Author: yeahpeng 4 * Created Time: 2015/6/4 22:03:06 5 * File Name: poj1724优先.cpp 6 * Description 7 8 N cities named with numbers 1 ... N are connected with one-way roads. Each road has two parameters associated with it : the road length and the toll that needs to be paid for the road (expressed in the number of coins). 9 Bob and Alice used to live in the city 1. After noticing that Alice was cheating in the card game they liked to play, Bob broke up with her and decided to move away - to the city N. He wants to get there as quickly as possible, but he is short on cash. 10 11 We want to help Bob to find the shortest path from the city 1 to the city N that he can afford with the amount of money he has. 12 Input 13 14 The first line of the input contains the integer K, 0 <= K <= 10000, maximum number of coins that Bob can spend on his way. 15 The second line contains the integer N, 2 <= N <= 100, the total number of cities. 16 17 The third line contains the integer R, 1 <= R <= 10000, the total number of roads. 18 19 Each of the following R lines describes one road by specifying integers S, D, L and T separated by single blank characters : 20 S is the source city, 1 <= S <= N 21 D is the destination city, 1 <= D <= N 22 L is the road length, 1 <= L <= 100 23 T is the toll (expressed in the number of coins), 0 <= T <=100 24 25 Notice that different roads may have the same source and destination cities. 26 Output 27 28 The first and the only line of the output should contain the total length of the shortest path from the city 1 to the city N whose total toll is less than or equal K coins. 29 If such path does not exist, only number -1 should be written to the output. 30 Sample Input 31 32 5 33 6 34 7 35 1 2 2 3 36 2 4 3 3 37 3 4 2 4 38 1 3 4 1 39 4 6 2 1 40 3 5 2 0 41 5 4 3 2 42 Sample Output 43 44 11 45 */ 46 #include <stdio.h> 47 #include <iostream> 48 #include <algorithm> 49 #include <sstream> 50 #include <stdlib.h> 51 #include <string.h> 52 #include <limits.h> 53 #include <vector> 54 #include <string> 55 #include <time.h> 56 #include <math.h> 57 #include <queue> 58 #include <stack> 59 #include <set> 60 #include <map> 61 #define INF 0x3f3f3f3f 62 #define Zero(x) memset((x),0, sizeof(x)) 63 #define Neg(x) memset((x), -1, sizeof(x)) 64 #define dg(x) cout << #x << " = " << x << endl 65 #define pk(x) push_back(x) 66 #define pok() pop_back() 67 #define eps 1e-8 68 #define pii pair<int, int> 69 #define pi acos(-1.0) 70 using namespace std; 71 typedef long long ll; 72 bool debug = true; 73 int OK = 1; 74 const int maxn = 110; 75 const int maxm = 10010; 76 int k, n, r; 77 struct node{ 78 int l, t, v; 79 node(int ll = 0, int tt = 0, int vv = 0){ 80 l = ll; 81 t = tt; 82 v = vv; 83 } 84 bool operator<(const node& a) const { 85 return l > a.l; 86 } 87 }; 88 vector<node> gra[maxn]; 89 int ans; 90 bool cal(){ 91 priority_queue<node> pq; 92 pq.push(node(0, 0, 1)); 93 node top,t; 94 int sz; 95 while(!pq.empty()){ 96 top = pq.top(); 97 pq.pop(); 98 if(top.v == n) { 99 ans = top.l; 100 return true; 101 } 102 sz = gra[top.v].size(); 103 for(int i = 0; i < sz; ++i){ 104 t = gra[top.v][i]; 105 if(t.t + top.t <= k) pq.push(node(t.l + top.l, t.t + top.t, t.v)); 106 } 107 108 } 109 return false; 110 } 111 int main(){ 112 //freopen("data.in","r",stdin); 113 //freopen("data.out","w",stdout); 114 //cin.sync_with_stdio(false); 115 int u, v, l, t; 116 while(scanf("%d", &k) != EOF){ 117 scanf("%d%d", &n, &r); 118 for(int i = 0; i <= n; ++i) gra[i].clear(); 119 for(int i = 0; i < r; ++i){ 120 scanf("%d%d%d%d", &u, &v, &l, &t); 121 gra[u].pk(node(l, t, v)); 122 } 123 if(cal()) cout << ans << endl; 124 else cout << -1 << endl; 125 } 126 return 0; 127 }
1 /* 2 ***********Good LUCK********** 3 * Author: yeahpeng 4 * Created Time: 2015/6/3 23:37:49 5 * File Name: hdu4544_优先.cpp 6 * 题意：Problem Description 7 湫湫减肥 8 越减越肥！ 9 10 最近，减肥失败的湫湫为发泄心中郁闷，在玩一个消灭免子的游戏。 11 游戏规则很简单，用箭杀死免子即可。 12 箭是一种消耗品，已知有M种不同类型的箭可以选择，并且每种箭都会对兔子造成伤害，对应的伤害值分别为Di（1 <= i <= M），每种箭需要一定的QQ币购买。 13 假设每种箭只能使用一次，每只免子也只能被射一次，请计算要消灭地图上的所有兔子最少需要的QQ币。 14 15 16 Input 17 输入数据有多组，每组数据有四行； 18 第一行有两个整数N，M（1 <= N, M <= 100000），分别表示兔子的个数和箭的种类； 19 第二行有N个正整数，分别表示兔子的血量Bi（1 <= i <= N）； 20 第三行有M个正整数，表示每把箭所能造成的伤害值Di（1 <= i <= M）； 21 第四行有M个正整数，表示每把箭需要花费的QQ币Pi（1 <= i <= M）。 22 23 特别说明： 24 1、当箭的伤害值大于等于兔子的血量时，就能将兔子杀死； 25 2、血量Bi，箭的伤害值Di，箭的价格Pi，均小于等于100000。 26 27 28 Output 29 如果不能杀死所有兔子，请输出”No”，否则，请输出最少的QQ币数，每组输出一行。 30 31 32 Sample Input 33 3 3 34 1 2 3 35 2 3 4 36 1 2 3 37 3 4 38 1 2 3 39 1 2 3 4 40 1 2 3 1 41 42 43 Sample Output 44 6 45 4 46 */ 47 #include <stdio.h> 48 #include <iostream> 49 #include <algorithm> 50 #include <sstream> 51 #include <stdlib.h> 52 #include <string.h> 53 #include <limits.h> 54 #include <vector> 55 #include <string> 56 #include <time.h> 57 #include <math.h> 58 #include <queue> 59 #include <stack> 60 #include <set> 61 #include <map> 62 #define INF 0x3f3f3f3f 63 #define Zero(x) memset((x),0, sizeof(x)) 64 #define Neg(x) memset((x), -1, sizeof(x)) 65 #define dg(x) cout << #x << " = " << x << endl 66 #define pk(x) push_back(x) 67 #define pok() pop_back() 68 #define eps 1e-8 69 #define pii pair<int, int> 70 #define pi acos(-1.0) 71 using namespace std; 72 typedef long long ll; 73 bool debug = false; 74 int OK = 1; 75 const int maxn = 100100; 76 int n, m; 77 ll ans; 78 int b[maxn]; 79 struct node{ 80 int cost, d; 81 node(int dd = 0, int cc = 0){ 82 d = dd; 83 cost = cc; 84 } 85 bool operator < (const node& a)const { 86 return d > a.d ; 87 } 88 }jj[maxn]; 89 90 priority_queue<int, vector<int>, greater<int> > qj; 91 bool cal(){ 92 int st = 0; 93 ans = 0; 94 for(int i = 0; i < n; ++i){ 95 while(st < m && jj[st].d >= b[i]){ 96 qj.push(jj[st].cost); 97 st++; 98 } 99 if(qj.empty()) return false; 100 if(debug){ 101 dg(qj.top()); 102 } 103 ans += qj.top(); 104 qj.pop(); 105 } 106 return true; 107 } 108 int main(){ 109 //freopen("data.in","r",stdin); 110 //freopen("data.out","w",stdout); 111 //cin.sync_with_stdio(false); 112 while(cin >> n >>m){ 113 while(!qj.empty()) qj.pop(); 114 for(int i = 0; i < n; ++i){ 115 scanf("%d", b + i); 116 } 117 sort(b, b + n, greater<int>()); 118 for(int i = 0; i < m; ++i){ 119 scanf("%d", &jj[i].d); 120 } 121 for(int i = 0; i < m; ++i){ 122 scanf("%d", &jj[i].cost); 123 } 124 sort(jj, jj + m); 125 if(cal()){ 126 cout << ans << endl; 127 }else cout << "No" << endl; 128 } 129 return 0; 130 }
1 /* 2 ***********Good LUCK********** 3 * Author: yeahpeng 4 * Created Time: 2015/6/3 23:09:57 5 * File Name: hdu1242_优队+bfs.cpp 6 */ 7 #include <stdio.h> 8 #include <iostream> 9 #include <algorithm> 10 #include <sstream> 11 #include <stdlib.h> 12 #include <string.h> 13 #include <limits.h> 14 #include <vector> 15 #include <string> 16 #include <time.h> 17 #include <math.h> 18 #include <queue> 19 #include <stack> 20 #include <set> 21 #include <map> 22 #define INF 0x3f3f3f3f 23 #define Zero(x) memset((x),0, sizeof(x)) 24 #define Neg(x) memset((x), -1, sizeof(x)) 25 #define dg(x) cout << #x << " = " << x << endl 26 #define pk(x) push_back(x) 27 #define pok() pop_back() 28 #define eps 1e-8 29 #define pii pair<int, int> 30 #define pi acos(-1.0) 31 using namespace std; 32 typedef long long ll; 33 bool debug = true; 34 int OK = 1; 35 int n, m; 36 const int maxn = 220; 37 char mp[maxn][maxn]; 38 struct node{ 39 int x, y, sp; 40 node(int xx = 0,int yy = 0, int ss = 0){ 41 x = xx; 42 y = yy; 43 sp = ss; 44 } 45 bool operator<(const node& b)const { 46 return sp > b.sp; 47 } 48 }st, ed; 49 int dir[4][2] = { 50 0, 1, 0, -1, 1, 0, -1, 0 51 }; 52 int ans; 53 bool flag[maxn][maxn]; 54 bool cal(){ 55 Zero(flag); 56 priority_queue<node> q; 57 q.push(node(st.x, st.y, 0)); 58 node top; 59 int x, y; 60 flag[st.x][st.y] = true; 61 while(!q.empty()){ 62 top = q.top(); 63 q.pop(); 64 if(top.x == ed.x && top.y == ed.y){ 65 ans = top.sp; 66 return true; 67 } 68 69 for(int i = 0; i < 4; ++i){ 70 x = top.x + dir[i][0]; 71 y = top.y + dir[i][1]; 72 if(!flag[x][y] && x >= 1 && x <= n && y >= 1 && y <= m){ 73 if(mp[x][y] != '#'){ 74 flag[x][y] = true; 75 if(mp[x][y] == 'x') q.push(node(x, y,top.sp + 2)); 76 else q.push(node(x, y, top.sp + 1)); 77 } 78 } 79 } 80 } 81 return false; 82 } 83 int main(){ 84 //freopen("data.in","r",stdin); 85 //freopen("data.out","w",stdout); 86 //cin.sync_with_stdio(false); 87 while(scanf("%d%d", &n, &m) != EOF){ 88 for(int i = 1; i <= n; ++i) scanf("%s", &mp[i][1]); 89 for(int i = 1; i <= n; ++i){ 90 for(int j = 1; j <= m; ++j){ 91 if(mp[i][j] == 'r'){ 92 st.x = i; 93 st.y = j; 94 }else if(mp[i][j] == 'a'){ 95 ed.x = i; 96 ed.y = j; 97 } 98 } 99 } 100 if(cal()){ 101 cout << ans << endl; 102 }else cout << "Poor ANGEL has to stay in the prison all his life." << endl; 103 } 104 return 0; 105 }
zoj 3632 watermelon full of watere
1 /***Good Luck***/ 2 #define _CRT_SECURE_NO_WARNINGS 3 #include <iostream> 4 #include <cstdio> 5 #include <cstdlib> 6 #include <cstring> 7 #include <string> 8 #include <algorithm> 9 #include <stack> 10 #include <map> 11 #include <queue> 12 #include <vector> 13 #include <set> 14 #include <functional> 15 #include <cmath> 16 #include <numeric> 17 18 #define Zero(a) memset(a, 0, sizeof(a)) 19 #define Neg(a) memset(a, -1, sizeof(a)) 20 #define All(a) a.begin(), a.end() 21 #define PB push_back 22 #define inf 0x3f3f3f3f 23 #define inf2 0x7fffffffffffffff 24 #define ll long long 25 using namespace std; 26 //#pragma comment(linker, "/STACK:102400000,102400000") 27 void get_val(int &a) { 28 int value = 0, s = 1; 29 char c; 30 while ((c = getchar()) == ' ' || c == '\n'); 31 if (c == '-') s = -s; else value = c - 48; 32 while ((c = getchar()) >= '0' && c <= '9') 33 value = value * 10 + c - 48; 34 a = s * value; 35 } 36 const int maxn = 50005; 37 int n; 38 ll arr1[maxn], arr2[maxn]; 39 ll dp[maxn]; 40 struct node { 41 ll v, i; 42 node(ll vv = 0, ll ii = 0) :i(ii), v(vv){} 43 bool operator<(const node& a) const { 44 return v > a.v || (v == a.v && i > a.i); 45 } 46 }; 47 int main() { 48 //freopen("data.out", "w", stdout); 49 //freopen("data.in", "r", stdin); 50 //cin.sync_with_stdio(false); 51 while (scanf("%d", &n) != EOF) { 52 for (int i = 1; i <= n; ++i) scanf("%lld", arr1 + i); 53 for (int i = 1; i <= n; ++i) scanf("%lld", arr2 + i); 54 priority_queue<node> pq; 55 dp[0] = 0; 56 for (int i = 1; i <= n; ++i) { 57 node tmp = node(dp[i - 1] + arr1[i], arr2[i] + i - 1); 58 while (!pq.empty() && pq.top().i < i) pq.pop(); 59 pq.push(tmp); 60 dp[i] = pq.top().v; 61 } 62 printf("%lld\n", dp[n]); 63 } 64 return 0; 65 }
hdu 5090
1 //************************* 2 //*******GL HF************ 3 //************************* 4 #include<cstdio> 5 #include<cmath> 6 #include<algorithm> 7 #include<iostream> 8 #include<cstring> 9 #include<cstdlib> 10 #include<map> 11 #include<queue> 12 #include<set> 13 #include<vector> 14 #pragma comment(linker, "/STACK:102400000,102400000") 15 #define Local 16 #define o(a) cout << #a << " = " << a << endl 17 #define pb(a) push_back(a) 18 #define lson l,mid,pos<<1 19 #define rson mid+1,r,pos<<1|1 20 typedef long long ll; 21 using namespace std; 22 23 void local(){ 24 #ifdef Local 25 freopen("date.in","r",stdin); 26 freopen("date.out","w",stdout); 27 #endif 28 } 29 const int maxn = 550; 30 bool debug = false; 31 int n, k; 32 int arr[maxn]; 33 priority_queue<int ,vector<int>,greater<int> > q; 34 int main(){ 35 int T; 36 scanf("%d", &T); 37 38 while(T--){ 39 scanf("%d%d", &n, &k); 40 int v; 41 while(!q.empty()) q.pop(); 42 for(int i = 1; i <= n ; ++i){ 43 scanf("%d", &v); 44 q.push(v); 45 } 46 int top; 47 int c = 1; 48 if(debug){ o(q.top());} 49 while(!q.empty() && q.top() <= n){ 50 top = q.top(); 51 q.pop(); 52 if(debug){ o(top);} 53 if(c != top){ 54 q.push(top + k); 55 }else { 56 c++; 57 } 58 59 //cout << "OK" << endl; 60 } 61 if(!q.empty()){ cout << "Tom" << endl;} 62 else cout << "Jerry" << endl; 63 } 64 }
