zoj 3229 Shoot the Bullet（有源汇上下界最大流）
Posted 日拱一卒 功不唐捐
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Gensokyo is a world which exists quietly beside ours, separated by a mystical border. It is a utopia where humans and other beings such as fairies, youkai(phantoms), and gods live peacefully together. Shameimaru Aya is a crow tengu with the ability to manipulate wind who has been in Gensokyo for over 1000 years. She runs the Bunbunmaru News - a newspaper chock-full of rumors, and owns the Bunkachou - her record of interesting observations for Bunbunmaru News articles and pictures of beautiful danmaku(barrange) or cute girls living in Gensokyo. She is the biggest connoisseur of rumors about the girls of Gensokyo among the tengu. Her intelligence gathering abilities are the best in Gensokyo!
During the coming n days, Aya is planning to take many photos of m cute girls living in Gensokyo to write Bunbunmaru News daily and record at least Gx photos of girl x in total in the Bunkachou. At the k-th day, there are Ck targets, Tk1, Tk2, ..., TkCk. The number of photos of target Tki that Aya takes should be in range [Lki, Rki], if less, Aya cannot write an interesting article, if more, the girl will become angry and use her last spell card to attack Aya. What‘s more, Aya cannot take more than Dk photos at the k-th day. Under these constraints, the more photos, the better.
Aya is not good at solving this complex problem. So she comes to you, an earthling, for help.
Input
There are about 40 cases. Process to the end of file.
Each case begins with two integers 1 <= n <= 365, 1 <= m <= 1000. Then m integers, G1, G2, ..., Gm in range [0, 10000]. Then n days. Each day begins with two integer 1 <= C <= 100, 0 <= D <= 30000. Then C different targets. Each target is described by three integers, 0 <= T < m, 0 <= L <= R <= 100.
Output
For each case, first output the number of photos Aya can take, -1 if it‘s impossible to satisfy her needing. If there is a best strategy, output the number of photos of each girl Aya should take at each day on separate lines. The output must be in the same order as the input. If there are more than one best strategy, any one will be OK.
Output a blank line after each case.
Sample Input
2 3 12 12 12 3 18 0 3 9 1 3 9 2 3 9 3 18 0 3 9 1 3 9 2 3 9 2 3 12 12 12 3 18 0 3 9 1 3 9 2 3 9 3 18 0 0 3 1 3 6 2 6 9 2 3 12 12 12 3 15 0 3 9 1 3 9 2 3 9 3 21 0 0 3 1 3 6 2 6 12
Sample Output
36 6 6 6 6 6 6 36 9 6 3 3 6 9 -1
题意：n天给m个女孩拍照，每个女孩至少要拍gi张，每一天要给c个不同的女孩拍照，每一天最多拍d张，每一天每个女孩最少拍L张，最多拍R张
问能否完成拍照，若能输出最多拍多少张，同时输出每一天每个女孩拍多少张
否则，输出-1
有源汇上下界最大流
源点向每一天连[0,d]的边，每个女孩向汇点连[gi,inf]的边，每一天向每个女孩连[L,R]的边
本题无法评测。。，所以代码不知对错
#include<cstdio> #include<algorithm> #include<cstring> #include<queue> #define N 1500 #define M 1500000 #define inf 2e9 using namespace std; int n,m,src,dec,S,T,sum; int tot,front[N],to[M],next[M],cap[M]; int lev[N],cur[N],a[N]; int ans[N],low[370][1001]; queue<int>q; void add(int u,int v,int w) { to[++tot]=v; next[tot]=front[u]; front[u]=tot; cap[tot]=w; to[++tot]=u; next[tot]=front[v]; front[v]=tot; cap[tot]=0; } bool bfs(int s,int t) { for(int i=0;i<=n+m+3;i++) cur[i]=front[i],lev[i]=-1; while(!q.empty()) q.pop(); q.push(s); lev[s]=0; int now; while(!q.empty()) { now=q.front(); q.pop(); for(int i=front[now];i;i=next[i]) if(cap[i]>0&&lev[to[i]]==-1) { lev[to[i]]=lev[now]+1; if(to[i]==t) return true; q.push(to[i]); } } return false; } int dfs(int now,int t,int flow) { if(now==t) return flow; int delta,rest=0; for(int & i=cur[now];i;i=next[i]) { if(lev[to[i]]>lev[now]&&cap[i]>0) { delta=dfs(to[i],t,min(flow-rest,cap[i])); if(delta) { cap[i]-=delta; cap[i^1]+=delta; rest+=delta; if(rest==flow) break; } } } if(rest!=flow) lev[now]=-1; return rest; } int dinic(int s,int t) { int tmp=0; while(bfs(s,t)) tmp+=dfs(s,t,inf); return tmp; } int main() { while(scanf("%d%d",&n,&m)!=EOF) { tot=1; sum=0; memset(a,0,sizeof(a)); memset(front,0,sizeof(front)); dec=n+m+1; S=n+m+2; T=n+m+3; add(dec,src,inf); int x; for(int i=1;i<=m;i++) { scanf("%d",&x); a[i]-=x; a[dec]+=x; add(i,dec,inf); } int c,d,y,z; for(int i=1;i<=n;i++) { scanf("%d%d",&c,&d); add(src,m+i,d); for(int j=1;j<=c;j++) { scanf("%d%d%d",&z,&x,&y); z++; add(m+i,z,y-x); low[i][z]=x; a[m+i]-=x; a[z]+=x; } } for(int i=0;i<=n+m+1;i++) ///i=0 if(a[i]<0) add(i,T,-a[i]); else if(a[i]>0) add(S,i,a[i]),sum+=a[i]; if(dinic(S,T)==sum) { //int okflow=cap[3]; cap[3]=-1; //printf("%d\n",dinic(src,dec)+okflow); dinic(src,dec); add(dec,src,-inf); int g=0; for(int i=front[src];i;i=next[i]) g+=cap[i^1]; printf("%d\n",g); for(int i=front[src];i;i=next[i]) { memset(ans,0,sizeof(ans)); if(to[i]<=m||to[i]>m+n) continue; for(int j=front[to[i]];j;j=next[j]) { if(to[j]>m) continue; ans[to[j]]=cap[j^1]; } for(int j=1;j<=m;j++) printf("%d\n",ans[j]+low[to[i]-m][j]); } } else printf("-1\n"); printf("\n"); } return 0; }
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