Mining Station on the Sea
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3565 Accepted Submission(s): 1108
Problem Description
The ocean is a treasure house of resources and the development of human society comes to depend more and more on it. In order to develop and utilize marine resources, it is necessary to build mining stations on the sea. However, due to seabed mineral resources, the radio signal in the sea is often so weak that not all the mining stations can carry out direct communication. However communication is indispensable, every two mining stations must be able to communicate with each other (either directly or through other one or more mining stations). To meet the need of transporting the exploited resources up to the land to get put into use, there build n ports correspondently along the coast and every port can communicate with one or more mining stations directly.
Due to the fact that some mining stations can not communicate with each other directly, for the safety of the navigation for ships, ships are only allowed to sail between mining stations which can communicate with each other directly.
The mining is arduous and people do this job need proper rest (that is, to allow the ship to return to the port). But what a coincidence! This time, n vessels for mining take their turns to take a rest at the same time. They are scattered in different stations and now they have to go back to the port, in addition, a port can only accommodate one vessel. Now all the vessels will start to return, how to choose their navigation routes to make the total sum of their sailing routes minimal.
Notice that once the ship entered the port, it will not come out!
Due to the fact that some mining stations can not communicate with each other directly, for the safety of the navigation for ships, ships are only allowed to sail between mining stations which can communicate with each other directly.
The mining is arduous and people do this job need proper rest (that is, to allow the ship to return to the port). But what a coincidence! This time, n vessels for mining take their turns to take a rest at the same time. They are scattered in different stations and now they have to go back to the port, in addition, a port can only accommodate one vessel. Now all the vessels will start to return, how to choose their navigation routes to make the total sum of their sailing routes minimal.
Notice that once the ship entered the port, it will not come out!
Input
There are several test cases. Every test case begins with four integers in one line, n (1 = <n <= 100), m (n <= m <= 200), k and p. n indicates n vessels and n ports, m indicates m mining stations, k indicates k edges, each edge corresponding to the link between a mining station and another one, p indicates p edges, each edge indicating the link between a port and a mining station. The following line is n integers, each one indicating one station that one vessel belongs to. Then there follows k lines, each line including 3 integers a, b and c, indicating the fact that there exists direct communication between mining stations a and b and the distance between them is c. Finally, there follows another p lines, each line including 3 integers d, e and f, indicating the fact that there exists direct communication between port d and mining station e and the distance between them is f. In addition, mining stations are represented by numbers from 1 to m, and ports 1 to n. Input is terminated by end of file.
Output
Each test case outputs the minimal total sum of their sailing routes.
Sample Input
3 5 5 6
1 2 4
1 3 3
1 4 4
1 5 5
2 5 3
2 4 3
1 1 5
1 5 3
2 5 3
2 4 6
3 1 4
3 2 2
Sample Output
13
Source
交了50多发 终于找出了最好的spfa板子。。。。
这是费用流做的
或者 求出每个船道港口的最短距离 在用hc就好了
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #include <bitset> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define rb(a) scanf("%lf", &a) #define rf(a) scanf("%f", &a) #define pd(a) printf("%d ", a) #define plld(a) printf("%lld ", a) #define pc(a) printf("%c ", a) #define ps(a) printf("%s ", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 300000, INF = 0x3f3f3f3f, LL_INF = 0x7fffffffffffffff; int n, m, k, q, s, t; int head[3000], d[3000], vis[3000], p[3000], f[3000], inc[3000], nex[maxn]; int flow, value, cnt; struct node { int u, v, w, c; }Node[maxn]; void add_(int u, int v, int w, int c) { Node[cnt].u = u; Node[cnt].v = v; Node[cnt].w = w; Node[cnt].c = c; nex[cnt] = head[u]; head[u] = cnt++; } void add(int u, int v, int w, int c) { add_(u, v, w, c); add_(v, u, -w, 0); } int spfa() { deque<int> Q; mem(vis, 0); mem(p, -1); mem(d, INF); d[s] = 0; Q.push_front(s); vis[s] = 1; p[s] = 0, f[s] = INF; while(!Q.empty()) { int u = Q.front(); Q.pop_front(); vis[u] = 0; for(int i = head[u]; i != -1; i = nex[i]) { node e = Node[i]; if(d[e.v] > d[u] + Node[i].w && Node[i].c > 0) { d[e.v] = d[u] + Node[i].w; p[e.v] = i; f[e.v] = min(f[u], Node[i].c); if(!vis[e.v]) { if(Q.empty()) Q.push_front(e.v); else { if(d[e.v] < d[Q.front()]) Q.push_front(e.v); else Q.push_back(e.v); } vis[e.v] = 1; } } } } if(p[t] == -1) return 0; flow += f[t]; value += f[t] * d[t]; for(int i = t; i != s; i = Node[p[i]].u) { Node[p[i]].c -= f[t]; Node[p[i] ^ 1].c += f[t]; } return 1; } void max_flow() { value = flow = 0; while(spfa()); pd(value); } void init() { mem(head, -1); cnt = 0; } int main() { while(scanf("%d%d%d%d", &n, &m, &k, &q) != EOF) { init(); s = 0; t = m + n + 1; int u, v, w, tmp; for(int i = 1; i <= n; i++) { add(i + m, t, 0, 1); scanf("%d", &tmp); add(s, tmp, 0, 1); } for(int i = 1; i <= k; i++) { scanf("%d%d%d", &u, &v, &w); add(u, v, w, INF); add(v, u, w, INF); } for(int i = 1; i <= q; i++) { scanf("%d%d%d", &u, &v, &w); add(v, m + u, w, 1); } max_flow(); } return 0; }
Mining Station on the Sea
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3565 Accepted Submission(s): 1108
Problem Description
The ocean is a treasure house of resources and the development of human society comes to depend more and more on it. In order to develop and utilize marine resources, it is necessary to build mining stations on the sea. However, due to seabed mineral resources, the radio signal in the sea is often so weak that not all the mining stations can carry out direct communication. However communication is indispensable, every two mining stations must be able to communicate with each other (either directly or through other one or more mining stations). To meet the need of transporting the exploited resources up to the land to get put into use, there build n ports correspondently along the coast and every port can communicate with one or more mining stations directly.
Due to the fact that some mining stations can not communicate with each other directly, for the safety of the navigation for ships, ships are only allowed to sail between mining stations which can communicate with each other directly.
The mining is arduous and people do this job need proper rest (that is, to allow the ship to return to the port). But what a coincidence! This time, n vessels for mining take their turns to take a rest at the same time. They are scattered in different stations and now they have to go back to the port, in addition, a port can only accommodate one vessel. Now all the vessels will start to return, how to choose their navigation routes to make the total sum of their sailing routes minimal.
Notice that once the ship entered the port, it will not come out!
Due to the fact that some mining stations can not communicate with each other directly, for the safety of the navigation for ships, ships are only allowed to sail between mining stations which can communicate with each other directly.
The mining is arduous and people do this job need proper rest (that is, to allow the ship to return to the port). But what a coincidence! This time, n vessels for mining take their turns to take a rest at the same time. They are scattered in different stations and now they have to go back to the port, in addition, a port can only accommodate one vessel. Now all the vessels will start to return, how to choose their navigation routes to make the total sum of their sailing routes minimal.
Notice that once the ship entered the port, it will not come out!
Input
There are several test cases. Every test case begins with four integers in one line, n (1 = <n <= 100), m (n <= m <= 200), k and p. n indicates n vessels and n ports, m indicates m mining stations, k indicates k edges, each edge corresponding to the link between a mining station and another one, p indicates p edges, each edge indicating the link between a port and a mining station. The following line is n integers, each one indicating one station that one vessel belongs to. Then there follows k lines, each line including 3 integers a, b and c, indicating the fact that there exists direct communication between mining stations a and b and the distance between them is c. Finally, there follows another p lines, each line including 3 integers d, e and f, indicating the fact that there exists direct communication between port d and mining station e and the distance between them is f. In addition, mining stations are represented by numbers from 1 to m, and ports 1 to n. Input is terminated by end of file.
Output
Each test case outputs the minimal total sum of their sailing routes.
Sample Input
3 5 5 6
1 2 4
1 3 3
1 4 4
1 5 5
2 5 3
2 4 3
1 1 5
1 5 3
2 5 3
2 4 6
3 1 4
3 2 2
Sample Output
13
Source