Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem F

Two Longest Paths
Input: Standard Input

Output: Standard Output

Tekuhp is a tourist city. There are N intersections in the city, connected with M one-way roads. Each one-way road connects from some intersection to another. There maybe many roads that connect a pair of intersections. To make the city very amazing, the roads are constructed so that it is not possible to start at some intersection, travel along the roads, and return to the starting intersection. (It remains a strange secret how people of Tekuhp return home from work each day.)

There are two groups of tourists planning to visit the city. They want to travel along the roads from some intersection to another. However, both groups do not want to run into each other. So they want two paths P₁ and P₂, each P_i , for 1 <= i <= 2, starts at some intersection s_i and ends at intersection t_i, such that both paths share no intersections, including the starting and the ending intersections. However, it is possible that a path P_i may contain only one node, i. e., s_i = t_i.

Tourists also want to visit many places. Since you are a good planner, you want to maximize the total number of intersections in both path.

Input

           First line of the input contains an integer T (1 <= T <= 10), the number of test cases. After that T test cases follow.

Each test case starts with integers N and M (1 <= N <= 300; 1 <= M <= 3,000), where N denotes the number of intersections and M denotes the number of roads. The intersections are numbered from 1 to N. After that M lines, describing road connection, follow. Each line contains two integers A and B denoting that there is a one-way road from intersection A to intersection B.

Output

         The output must contain T lines, each line for each test case. For each test case, the output contains an integer L denoting the maximum number of intersections in two non-intersecting paths.

Problemsetter: Jittat Fakcharoenphol

題目描述：

給一個有向無環的圖，求任意起點終點的兩條不相交的路徑，路徑上的點越多越好。

題目解法：

既然無環，可以使用最少費用流求解，多設兩個虛點到所有點的流量是 2。

由於每個點都只能用 1 次，因此要將每個點拆成入點和出點，中間的流量是 1。

#include <stdio.h>
#include <string.h>
#include <queue>
#include <deque>
using namespace std;
struct Node {
    int x, y, cap, cost;// x->y, v
    int next;
} edge[1000005];
int e, head[10005], dis[10005], prev[10005], record[10005], inq[10005];
void addEdge(int x, int y, int cap, int cost) {
    edge[e].x = x, edge[e].y = y, edge[e].cap = cap, edge[e].cost = cost;
    edge[e].next = head[x], head[x] = e++;
    edge[e].x = y, edge[e].y = x, edge[e].cap = 0, edge[e].cost = -cost;
    edge[e].next = head[y], head[y] = e++;
}
int mincost(int s, int t) {
    int mncost = 0, flow, totflow = 0;
    int i, x, y;
    while(1) {
        memset(dis, 63, sizeof(dis));
        int oo = dis[0];
        dis[s] = 0;
        deque<int> Q;
        Q.push_front(s);
        while(!Q.empty()) {
            x = Q.front(), Q.pop_front();
            inq[x] = 0;
            for(i = head[x]; i != -1; i = edge[i].next) {
                y = edge[i].y;
                if(edge[i].cap > 0 && dis[y] > dis[x] + edge[i].cost) {
                    dis[y] = dis[x] + edge[i].cost;
                    prev[y] = x, record[y] = i;
                    if(inq[y] == 0) {
                        inq[y] = 1;
                        if(Q.size() && dis[Q.front()] > dis[y])
                            Q.push_front(y);
                        else
                            Q.push_back(y);
                    }
                }
            }
        }
        if(dis[t] == oo)
            break;
        flow = oo;
        for(x = t; x != s; x = prev[x]) {
            int ri = record[x];
            flow = min(flow, edge[ri].cap);
        }
        for(x = t; x != s; x = prev[x]) {
            int ri = record[x];
            edge[ri].cap -= flow;
            edge[ri^1].cap += flow;
            edge[ri^1].cost = -edge[ri].cost;
        }
        totflow += flow;
        mncost += dis[t] * flow;
    }
    return mncost;
}
int main() {
    int testcase;
    int n, m, x, y;
    int i, j, k;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d", &n, &m);
        memset(head, -1, sizeof(head));
        e = 0;
        int st = 2 * n, ed = 2 * n + 1;
        int t1 = 2 * n + 2, t2 = 2 * n + 3;
        for(i = 0; i < n; i++) {
            addEdge(2 * i, 2 * i + 1, 1, -1);
            addEdge(t1, 2 * i, 1, 0);
            addEdge(2 * i + 1, t2, 1, 0);
        }
        addEdge(st, t1, 2, 0);
        addEdge(t2, ed, 2, 0);
        for(i = 0; i < m; i++) {
            scanf("%d %d", &x, &y);
            x--, y--;
            addEdge(2 * x + 1, 2 * y, 1, 0);
        }
        printf("%d\n", -mincost(st, ed));
    }
    return 0;
}