bzoj3280 小R的烦恼（费用流，类似餐巾计划）

把每一天拆成两个点，像餐巾计划那题一样的建图，跑最小费用流就可以了。

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define N 110inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();    return x*f;}int n,m,K,h[N],num=1,dis[N],path[N],T=107,s=101,tot=0;bool inq[N];struct edge{    int to,next,w,c;}data[6000];inline void add(int x,int y,int w,int c){    data[++num].to=y;data[num].next=h[x];h[x]=num;data[num].w=w;data[num].c=c;    data[++num].to=x;data[num].next=h[y];h[y]=num;data[num].w=0;data[num].c=-c;}inline bool spfa(){    deque<int>q;memset(dis,inf,sizeof(dis));memset(path,0,sizeof(path));    q.push_back(0);inq[0]=1;dis[0]=0;    while(!q.empty()){        int x=q.front();q.pop_front();inq[x]=0;        for(int i=h[x];i;i=data[i].next){            int y=data[i].to;if(!data[i].w) continue;            if(dis[x]+data[i].c<dis[y]){                dis[y]=dis[x]+data[i].c;path[y]=i;                if(!inq[y]){                    if(!q.empty()&&dis[y]<dis[q.front()]) q.push_front(y);                    else q.push_back(y);inq[y]=1;                }            }        }    }return path[T];}int main(){//  freopen("a.in","r",stdin);    int tst=read();    for(int owo=1;owo<=tst;++owo){        n=read();m=read();K=read();num=1;memset(h,0,sizeof(h));tot=0;        int ans=0,mxflow=0;printf("Case %d: ",owo);        for(int i=1;i<=n;++i){            int x=read();tot+=x;add(0,i,x,0);add(i+n,T,x,0);        }for(int i=1;i<=m;++i){            int x=read();add(0,s,x,read());        }for(int i=1;i<=K;++i){            int x=read(),val=read();            for(int j=1;j+x+1<=n;++j) add(j,j+x+1+n,inf,val);        }for(int i=1;i<n;++i) add(i,i+1,inf,0),add(s,i+n,inf,0);add(s,n+n,inf,0);        while(spfa()){            int low=inf,now=T;            while(path[now]) low=min(low,data[path[now]].w),now=data[path[now]^1].to;            mxflow+=low;ans+=low*dis[T];now=T;            while(path[now]) data[path[now]].w-=low,data[path[now]^1].w+=low,now=data[path[now]^1].to;        }if(mxflow!=tot) puts("impossible");        else printf("%d\n",ans);    }return 0;}