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; }