BZOJ2750: [HAOI2012]Road

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2750

跑n遍spfa，然后对于一条边，我们记录有多少最短路可以走到出点，记为a，从入点出发有多少最短路，记为b，那么每次就加上a[u]*b[v]就可以了。

（数据范围开错傻逼好久。。。药丸

#include<cstring>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define mm 1000000007
#define ll long long
#define maxm 5050
#define maxn 2050
using namespace std;
struct data{int from,obj,pre,c;
}e[maxm];
int tot,n,m;
int head[maxn],vis[maxn],dis[maxn],d[maxn];
ll ans[maxm],a[maxn],b[maxn];
int read(){
int x=0,f=1; char ch=getchar();
while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while (isdigit(ch)) {x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
void insert(int x,int y,int z){
e[++tot].obj=y; e[tot].from=x;  e[tot].c=z; e[tot].pre=head[x]; head[x]=tot;
}
void spfa(int s){
queue<int >q; clr(dis,127/3); clr(vis,0); dis[s]=0;
q.push(s);
while (!q.empty()){
int u=q.front(); q.pop(); vis[u]=1;
for (int j=head[u];j;j=e[j].pre){
int v=e[j].obj;
if (dis[v]>dis[u]+e[j].c){
dis[v]=dis[u]+e[j].c;
if (!vis[v]) {
vis[v]=1; q.push(v);
}
}
}
vis[u]=0;
}
}
void dfs(int u){
for (int j=head[u];j;j=e[j].pre){
int v=e[j].obj;
if (dis[v]==dis[u]+e[j].c){
a[v]=(a[v]+a[u])%mm;
d[v]--;
if (!d[v])dfs(v);
}
}
}
void dfs2(int u){
b[u]=1;
for (int j=head[u];j;j=e[j].pre){
int v=e[j].obj;
if (dis[v]==dis[u]+e[j].c){
if (!b[v]) dfs2(v);
b[u]=(b[u]+b[v])%mm;
}
}
}
void get(int u){
vis[u]=1;
for (int j=head[u];j;j=e[j].pre){
int v=e[j].obj;
if (dis[v]==dis[u]+e[j].c){
d[v]++;
if (!vis[v]) get(v);
}
}
}
int main(){
n=read(); m=read();
int x,y,z;
rep(i,1,m){
x=read(); y=read(); z=read();
insert(x,y,z);
}
rep(i,1,n){
spfa(i); clr(a,0); clr(b,0); clr(vis,0); clr(d,0);
get(i);
a[i]=1; dfs(i);
dfs2(i);
rep(j,1,m) if (dis[e[j].from]+e[j].c==dis[e[j].obj]) ans[j]=(ans[j]+a[e[j].from]*b[e[j].obj])%mm;
}
rep(i,1,m) printf("%lld\n",ans[i]);
return 0;
}