DP & graph theory DAY 5 am

POJ 1125 Stockbroker Grapevine

There are N number stockbroker can pass messages to each other, between them there are some one-way
communication path. Now there is a message to be delivered by a person start to all other
men, and should be delivered by which individuals can in the shortest time for all to take
receipt of the message.

Solution

Global Shortest

Floyd

POJ 1502 MPI Maelstrom

Given N number of information transmission between a lower triangular matrix, the time required for the processor, request information
transmitted to all other processors from the first processor time required maximum.

Solution

Single Source longest path dij (nm logn)

n ^ 2 logn

POJ 1511 Invitation Cards

N points of M have edges to FIG asking for all i , from the point a starting point i and
the return point 1 the time required minimum.

Solution

# Forward build Figure + single-source shortest
# reverse the construction diagram + single-source shortest, it is to have the edge in turn, starting from point 1 dij actually reverse shortest path

POJ 1724 ROADS

There are N cities and M unidirectional road, each road having a length and two charge properties.
In seeking the total cost does not exceed K premise from city 1 to the city N Shortest Path.

Solution

Dijkstra , while maintaining the current path of the total cost, cost more than the upper limit of the state no longer transfer

For each point split into k points, (u, 0), (u, 1), ....... (u, k)

Layered graph

POJ 1797 Heavy Transportation

Gives N points M no to FIG edges, each edge has a maximum load.
Request from point 1 to point N maximum weight can pass.

Solution

1. bipartite answer, assume that the maximum load> = mid, only by limiting the weight> = mid side, added to the figure, the connectivity is determined

2. KRUS maximum spanning tree, 1 --- n connectivity is determined, the communication is terminated

Descending plus side, the first 1 and n connectivity so that edge is the answer

Disjoint-set maintenance before adding k connectivity when large side

3.dijkstra variant single-source shortest path length to the right side of the minimum path

if ( dis[v] < min(dis[u],w) ) dis[v] = min( dis[u] , w )

POJ 1062 costly dowry ( http://poj.org/problem?id=1062)

PS: Note that chieftaincy is not necessarily the highest

Solution

For the first position in ascending order, and then enumerate [Emirates, Emirates + m] of this interval as the maximum position

Running time interval length of the shortest path

Dij ran away from super original point, and then to point 1

1. FIG construction, the article as a point, as substitutes side length pricing

Such as chiefs daughter <------ crystal ball weights are discounted prices

From an article exchange is to promise a path Emirates

Irrespective of the status of limit, starting from the chiefs promised to seek a single source shortest path to know a little from any desired gold

This is a reverse shortest reverse dij

2. Enumeration grade range position, the point is not within the range not through the level position, enum O ( N )
for "indirect transaction" method, a difference of about enumeration endpoint hierarchy interval <= m

I.e. max - min <= m, in order to make more room in a transaction section, we set the max-min = m

Then consider this transaction can be inside the range, it is illegal in the figure deleted

See the data, m may be large, but small n, then consider enumerated as the section end point

Figure 3. The remaining legitimate reverse run dij

BZOJ 3040 Shortest *

N 个点， M 条边的有向图，求点 1 到点 N 的最短路（保证存在）。
1 ≤ N ≤ 1000000, 1 ≤ M ≤ 10000000
By lydrainbowcat
边集分为两部分：
# 随机生成
# 输入给出

Solution

1. 采用高效的堆来优化 Dijkstra 算法。
# 斐波那契堆
# 配对堆

dij ，一个大图就凉

SPFA太小，凉

dij 用支持修改删除的堆，配对堆

2.用 pq 做，随机生成的数据不要，直接删掉

https://paste.ubuntu.com/p/Mp2fXMKv8J/

*d就是只输入不赋值

priority_queue STL实现合并

启发式合并，两个堆，A，B，把较小的堆拆了，放到较大的堆里面

每个点最多产生 logn 次代价

最小生成树算法

Prim

// Prim

void prim() {
    memset(dis, inf, sizeof dis);
    heap.push(data(1, dis[1] = 0));
    int ans = 0;
    while (!heap.empty()) {
        data t = heap.top(); heap.pop();
        if (dis[t.u] != t.d)continue;
        ans += t.d;
        for (int i = hd[t.u], v, w; i; i = nt[i])
            if (v = to[i], w = vl[i], dis[v] > w)
                heap.push(data(v, dis[v] = w));
    }
}