Marriage Match IV【最短路+最大流】

HDU-3416

  就是问，每条边只能走过一次的前提，并且要求每次走到终点的路都要是最短路，那么，问题就是我们先把所有的最短路的边给求出来，在这些边上去跑一个边权为1的最大流就可以了。

[code]#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 1e3 + 7, maxE = 2e5 + 7;
int N, M, head[maxN], cur[maxN], cnt, st, ed, top[2][maxN], tot[2];
struct Path
{
int nex, u, to, val;
Path(int a=-1, int me = 0, int b=0, int c=0):nex(a), u(me), to(b), val(c) {}
}act[2][maxE];
inline void TO(int u, int v, int w, int op)
{
act[op][tot[op]] = Path(top[op][u], u, v, w);
top[op][u] = tot[op]++;
}
struct Eddge
{
int nex, to, flow;
Eddge(int a=-1, int b=0, int c=0):nex(a), to(b), flow(c) {}
}edge[maxE];
inline void addEddge(int u, int v, int w)
{
edge[cnt] = Eddge(head[u], v, w);
head[u] = cnt++;
}
inline void _add(int u, int v, int w) { addEddge(u, v, w); addEddge(v, u, 0); }
int deep[maxN], dist[2][maxN];
queue<int> Q;
bool inque[maxN];
inline void spfa(int beg, int fin, int op)
{
while(!Q.empty()) Q.pop();
for(int i=1; i<=N; i++) dist[op][i] = INF;
dist[op][beg] = 0;
Q.push(beg);
while(!Q.empty())
{
int u = Q.front(); Q.pop(); inque[u] = false;
for(int i=top[op][u], v, w; ~i; i=act[op][i].nex)
{
v = act[op][i].to; w = act[op][i].val;
if(dist[op][v] > dist[op][u] + w)
{
dist[op][v] = dist[op][u] + w;
if(!inque[v])
{
Q.push(v);
inque[v] = true;
}
}
}
}
}
inline bool bfs()
{
for(int i=1; i<=N; i++) deep[i] = 0;
while(!Q.empty()) Q.pop();
Q.push(st); deep[st] = 1;
while(!Q.empty())
{
int u = Q.front(); Q.pop();
for(int i=head[u], v, f; ~i; i=edge[i].nex)
{
v = edge[i].to; f = edge[i].flow;
if(f && !deep[v])
{
deep[v] = deep[u] + 1;
Q.push(v);
}
}
}
return deep[ed];
}
inline int dfs(int u, int dist)
{
if(u == ed) return dist;
for(int &i=cur[u], v, f; ~i; i=edge[i].nex)
{
v = edge[i].to; f = edge[i].flow;
if(f && deep[v] == deep[u] + 1)
{
int di = dfs(v, min(dist, f));
if(di)
{
edge[i].flow -= di;
edge[i^1].flow += di;
return di;
}
}
}
return 0;
}
inline int Dinic()
{
int ans = 0, tmp;
while(bfs())
{
for(int i=1; i<=N; i++) cur[i] = head[i];
while((tmp = dfs(st, INF))) ans += tmp;
}
return ans;
}

inline void init()
{
cnt = tot[0] = tot[1] = 0;
for(int i=0; i<=N; i++) head[i] = top[0][i] = top[1][i] = -1;
}
int main()
{
int T; scanf("%d", &T);
while(T--)
{
scanf("%d%d", &N, &M);
init();
for(int i=1, u, v, w; i<=M; i++)
{
scanf("%d%d%d", &u, &v, &w);
if(u == v) continue;
TO(u, v, w, 0); //0去
TO(v, u, w, 1); //1回
}
scanf("%d%d", &st, &ed);
spfa(st, ed, 0); spfa(ed, st, 1);
int minn_load = dist[0][ed];
if(minn_load == INF) { printf("0\n"); continue; }
for(int i=0, u, v; i<tot[0]; i++)
{
u = act[0][i].u; v = act[0][i].to;
if(dist[0][u] + dist[1][v] + act[0][i].val == minn_load)
{
_add(u, v, 1);
}
}
printf("%d\n", Dinic());
}
return 0;
}