uva 11823 - Two Longest Paths （最小费用流）
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 Pi , for 1 <= i <= 2, starts at some intersection si and ends at intersection ti, such that both paths share no intersections, including the starting and the ending intersections. However, it is possible that a path Pi may contain only one node, i. e., si = ti.
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 Mdenotes 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.
| Sample Input |
Output for Sample Input |
| 3 3 2 1 2 2 3 8 9 1 2 2 3 3 4 5 6 6 7 7 8 2 6 6 3 3 7 4 2 1 2 3 4 |
3 8 4 |
Problemsetter: Jittat Fakcharoenphol
找到两条没有任何交点的路径，使它们经过的点数最多。每个点拆点，容量为1，费用-1，求流量为2的最小费用流。
//11823
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 602 + 10;
const int INF = 1000000000;
struct Edge {
int from, to, cap, flow, cost;
};
struct MCMF {
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn]; // 是否在队列中
int d[maxn]; // Bellman-Ford,单位流量的费用
int p[maxn]; // 上一条弧
int a[maxn]; // 可改进量
void init(int n) {
this->n = n;
for(int i = 0; i < n; i++) G[i].clear();
edges.clear();
}
void AddEdge(int from, int to, int cap, int cost) {
edges.push_back((Edge){from, to, cap, 0, cost});
edges.push_back((Edge){to, from, 0, 0, -cost});
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool BellmanFord(int s, int t, int &flow,int &cost) {
for(int i = 0; i < n; i++) d[i] = INF;
memset(inq, 0, sizeof(inq));
d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
queue<int> Q;
Q.push(s);
while(!Q.empty()) {
int u = Q.front(); Q.pop();
inq[u] = 0;
for(int i = 0; i < G[u].size(); i++) {
Edge& e = edges[G[u][i]];
if(e.cap > e.flow && d[e.to] > d[u] + e.cost) {
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
}
}
}
if(d[t] == INF) return false;//s-t不连通，失败退出
flow += a[t];
cost += d[t] * a[t];
int u = t;
while(u != s) {
edges[p[u]].flow += a[t];
edges[p[u]^1].flow -= a[t];
u = edges[p[u]].from;
}
return true;
}
// 需要保证初始网络中没有负权圈
int Mincost(int s, int t) {
int flow = 0,cost = 0;
while(BellmanFord(s, t,flow, cost));
return cost;
}
};
MCMF g;
int main(){
int t;
scanf("%d", &t);
while(t--){
int n, m;
scanf("%d%d", &n, &m);
int source = 0, sink1 = 2*n+1, sink2 = 2*n+2;
g.init(2*n+3);
while(m--){
int x, y;
scanf("%d%d", &x, &y);
g.AddEdge(x+n, y, 1, 0);
}
for(int i = 1;i <= n;i++){
g.AddEdge(source, i, 1, 0);
g.AddEdge(i+n, sink1, 1, 0);
g.AddEdge(i, i+n, 1, -1);
}
g.AddEdge(sink1, sink2, 2, 0);
int ans = -g.Mincost(source, sink2);
printf("%d\n", ans);
}
}