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;
}