POJ 3463 Sightseeing （最短路&次短路条数问题）

题意：给一个有向图，求从s到f 的最短路+最短路-1的条数。   有重边。

分析： 代码是根据挑战的次短路改编的。      具体请看代码。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<vector>
#include<cctype>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<iomanip>
#include<sstream>
#include<limits>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const ll INF = 1e18;
const int maxn = 2e5+10;
const ll MOD = 1000000007;
const double EPS = 1e-10;
const double Pi = acos(-1.0);

struct edge{int to,cost;};
//typedef pair<int,int>P; //1 dist 2 u
struct P{
int first,second,z; //z 标志是最短路还是次短路
P(int a=0,int b=0,int c=0):first(a),second(b),z(c){}
bool operator < (const P &rhs)const{return first>rhs.first;}
};
int V;
vector<edge>G[maxn];
int d[maxn],d2[maxn],cnt1[maxn],cnt2[maxn],vis[maxn][2]; //d cnt1 最短路长度及数量 d2 cnt2 次短路长度及数量 vis代表有没访问过
void dij(int s)
{
priority_queue<P >que;
fill(d,d+V+1,inf);
fill(d2,d2+V+1,inf);
d[s] = 0;
cnt1[s] = 1;
que.push(P(0,s,0));
while(!que.empty())
{
P p = que.top(); que.pop();
int v = p.second, z = p.z;
if (d2[v] < p.first || vis[v][z]) continue;
vis[v][z] = 1; //一定要记录是否访问过，不然会出错，因为是单调队列。
for(int i = 0; i < G[v].size();i++)
{
edge e = G[v][i];
int temp = p.first + e.cost;

if (d[e.to] > temp) // 比最短路短
{
d2[e.to] = d[e.to];
d[e.to] = temp;
cnt2[e.to] = cnt1[e.to];
cnt1[e.to] = z==0? cnt1[v]:cnt2[v]; //要弄清楚等于哪个。

que.push(P(d[e.to] ,e.to,0));
que.push(P(d2[e.to] ,e.to,1));
}else
if (d[e.to] == temp) cnt1[e.to] += z==0? cnt1[v]:cnt2[v]; // 等于最短路
else
if (temp < d2[e.to]) // 比次短路短
{
d2[e.to] = temp;
cnt2[e.to] = z==0? cnt1[v]:cnt2[v];
que.push(P(d2[e.to] ,e.to,1));
}else if (temp == d2[e.to]) cnt2[e.to] += z==0? cnt1[v]:cnt2[v]; // 等于次短路
}
}
}
int main(){
#ifdef LOCAL
freopen("C:\\Users\\lanjiaming\\Desktop\\acm\\in.txt","r",stdin);
//freopen("output.txt","w",stdout);
#endif
//ios_base::sync_with_stdio(0);
int T,n,m;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(int i = 0; i <= n; i++) G[i].clear();
V = n;
for(int i = 0; i < m; i++)
{
int a,b,l;
scanf("%d%d%d",&a,&b,&l);
G[a].push_back(edge{b,l});
}
int s,f;
scanf("%d%d",&s,&f);
memset(cnt1,0,sizeof cnt1);
memset(cnt2,0,sizeof cnt2);
memset(vis,0,sizeof vis);
dij(s);

if (d[f] == d2[f]-1) cnt1[f] += cnt2[f];
printf("%d\n",cnt1[f]);
}
return 0;
}