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跟着金主席的步伐，一步两步，一步两步，一步两步似poi~

嘿嘿嘿， 看着 金主席做$poi$，然后看了看我以前做过的$poi$，想想把能做的做完好辣

现在做了 几道：   19/20

$4.6$ $qiancl:$一半辣一半辣

$4.7$ $qiancl:$拿了一血好开心啊

-->http://qiancl.is-programmer.com/user_files/qiancl/Image/Problem%203072.%20--%20[Pa2012]Two%20Cakes.html

……第一眼挺简单-。-

……第二眼不会-。-|||

……考虑DP$f[i][j]$表示 $a$串取了$i$次，$b$串取了$j$次，于是

$f[i-1][j] -> f[i][j]$

$f[i][j-1] -> f[i][j]$

$f[i-1][j-1] -> f[i][j] (a[i]!=b[j])$

#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define nc() getchar()
inline int read(){
    int x=0;char ch=nc();for(;ch<'0'||ch>'9';ch=nc());
    for(;ch<='9'&&ch>='0';x=x*10+ch-48,ch=nc());return x;
}
#define N 1000100
int i,j,k,m,n,x,y,a[N],b[N],rk[N],f[5005][5005],g[N];
int main(){
    for(n=read(),i=1;i<=n;++i)a[i]=read(),rk[a[i]]=i;
    for(i=1;i<=n;++i)b[i]=read();
    memset(f,0x3f,sizeof f);
    f[0][0]=0;
    for(i=1;i<=n;++i)
        for(j=1;j<=n;++j){
            f[i][j]=min(f[i][j],f[i-1][j]+1);
            f[i][j]=min(f[i][j],f[i][j-1]+1);
            if(a[i]!=b[j])f[i][j]=min(f[i][j],f[i-1][j-1]+1);
        }
    printf("%d\n",f[n][n]);
}

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

不说什么了，n遍spfa然后Dp

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
#define nc() getchar()
inline int read(){
	int x=0;char ch=nc();for(;ch<'0'||ch>'9';ch=nc());
	for(;ch<='9'&&ch>='0';x=x*10+ch-48,ch=nc());return x;
}
#define N 1550
#define ll long long
int i,j,k,m,n,x,y,cnt,last[N],dis[N],d[N],a[N],b[N];ll ans[50050];
struct edge{int from,to,next,v;}e[50050];bool vis[N];
inline void add(int w,int v,int u){e[++cnt]=(edge){u,v,last[u],w},last[u]=cnt;}
#define met(x,y) memset(x,y,sizeof x)
inline void spfa(int s){
	queue<int>q;met(dis,0x3f),met(vis,0),dis[s]=0,q.push(s);
	while(!q.empty()){
		int k=q.front();q.pop(),vis[k]=1;
		for(int i=last[k],y;i;i=e[i].next)if(dis[y=e[i].to]>dis[k]+e[i].v){
			dis[y]=dis[k]+e[i].v;
			if(!vis[y])vis[y]=1,q.push(y);
		}vis[k]=0;
	}
}
#define mo 1000000007
inline void ad(int&a,int b){a+=b;if(a>=mo)a-=mo;}
inline void ad(ll&a,ll b){a+=b;if(a>=mo)a-=mo;}
void dfs1(int x){vis[x]=1;for(int i=last[x],y;i;i=e[i].next)if(dis[y=e[i].to]==dis[x]+e[i].v){++d[y];if(!vis[y])dfs1(y);}}
void dfs2(int x){for(int i=last[x],y;i;i=e[i].next)if(dis[y=e[i].to]==dis[x]+e[i].v){ad(a[y],a[x]);if(!(--d[y]))dfs2(y);}}
void dfs3(int x){b[x]=1;for(int i=last[x],y;i;i=e[i].next)if(dis[y=e[i].to]==dis[x]+e[i].v){if(!b[y])dfs3(y);ad(b[x],b[y]);}}
int main(){
	for(n=read(),m=read(),i=1;i<=m;++i)add(read(),read(),read());
	for(i=1;i<=n;++i){
		spfa(i),met(vis,0),met(a,0),met(b,0),met(d,0);
		dfs1(i),a[i]=1,dfs2(i),dfs3(i);
		for(j=1;j<=m;++j)if(dis[e[j].from]+e[j].v==dis[e[j].to])ad(ans[j],1ll*a[e[j].from]*b[e[j].to]);
	}
	for(i=1;i<=m;++i)printf("%lld\n",ans[i]);
}