【zoj2760】【最大流】How Many Shortest Path

求一个带权有向图中s-t边不相交最短路最多有几条。

建图方法：分别从源点和汇点作一次dijkstra，然后建一个满足ds[u] + w[u][v] + dt[v] == ds[t]这样的关系的导出子图，这样做是为了保证网络图中任意一条s-t的路都是最短路，每条边的容量都为1，然后做一次最大流就行了。

另外这道题比较坑的地方是：矩阵对角线上的长度不一定为0。。。

代码:

#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 100 + 10;
int n,s,t;
int ds[maxn],dt[maxn];
int map[maxn][maxn],tmp[maxn][maxn];
int dis[maxn],gap[maxn],pre[maxn];
int cur[maxn],cap[maxn][maxn];
bool done[maxn];

void init()
{
freopen("zoj2760.in","r",stdin);
freopen("zoj2760.out","w",stdout);
}

inline int min(int a,int b)
{
return a < b ? a : b;
}

void dijkstra(int st,int op)
{
typedef pair<int ,int>pii;
priority_queue<pii,vector<pii>,greater<pii> >q;
memset(done,false,sizeof(done));
int *dis;
if(op == 0)dis = ds;
else dis = dt;
q.push(make_pair(dis[st] = 0,st));
while(!q.empty())
{
pii u = q.top();q.pop();
if(done[u.second])continue;
int k = u.second;
done[k] = true;
for(int i = 0;i < n;i++)
{
if(done[i] || map[k][i] == inf)continue;
if(dis[k] + map[k][i] < dis[i])
{
dis[i] = dis[k] + map[k][i];
q.push(make_pair(dis[i],i));
}
}
}
}

void make_map()
{
for(int u = 0;u < n;u++)
{
for(int v = 0;v < n;v++)
{
if(u == v || ds[u] == inf || dt[v] == inf)continue;
if(map[v][u] != inf && ds[u] + map[v][u] + dt[v] == ds[t])
cap[u][v] = 1;
}
}
}

int sap()
{
memset(pre,0,sizeof(pre));
memset(dis,0,sizeof(dis));
memset(gap,0,sizeof(gap));
memset(cur,0,sizeof(cur));
int u = pre[s] = s,maxflow = 0,aug = inf;
gap[s] = n;
while(dis[s] < n)
{
loop: for(int v = cur[u];v < n;v++)
{
if(cap[u][v] && dis[u] == dis[v] + 1)
{
cur[u] = v;
aug = min(aug,cap[u][v]);
pre[v] = u;
u = v;
if(v == t)
{
maxflow += aug;
for(u = pre[u];v != s;v = u,u = pre[u])
{
cap[u][v] -= aug;
cap[v][u] += aug;
}
aug = inf;
}
goto loop;
}
}
int mind = n;
for(int v = 0;v < n;v++)
{
if(cap[u][v] && (mind > dis[v]))
{
cur[u] = v;
mind = dis[v];
}
}
if((--gap[dis[u]]) == 0)break;
gap[dis[u] = mind + 1]++;
u = pre[u];
}
return maxflow;
}

void reverse()
{
for(int i = 0;i < n;i++)
{
for(int j = 0;j < n;j++)
{
tmp[j][i] = map[i][j];
}
}
for(int i = 0;i < n;i++)
{
for(int j = 0;j < n;j++)
{
map[i][j] = tmp[i][j];
}
}
}

void solve()
{
memset(cap,0,sizeof(cap));
memset(ds,0x3f,sizeof(ds));
memset(dt,0x3f,sizeof(dt));
dijkstra(s,0);
reverse();
dijkstra(t,1);
make_map();
printf("%d\n",sap());
}

void readdata()
{
while(~scanf("%d",&n))
{
memset(map,0,sizeof(map));
for(int i = 0;i < n;i++)
{
for(int j = 0;j < n;j++)
{
scanf("%d",&map[i][j]);
if(i == j)map[i][j] = 0;
if(map[i][j] == -1)map[i][j] = inf;
}
}
scanf("%d%d",&s,&t);
if(s == t)
{
printf("inf\n");
continue;
}
solve();
}
}

int main()
{
init();
readdata();
return 0;
}