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
图论 网络流
复习了一波有上下界网络流的姿势。
顺便复习了一下东方的设定
题面里的罗马音居然都能看懂，是该说新标日姿势没忘光呢还是四斋蒸鹅心呢
如何求带上下界的最大流？
在求出可行流以后，删掉超级源汇，从旧源到旧汇跑网络流，把能用的残余流量都流掉，再加上原先的下界即可。
好像这道题的SPJ挂掉了，近几个月的提交结果全是Judge Internal Error
代码不保证正确，仅供参考
1 /*by SilverN*/ 2 #include<iostream> 3 #include<algorithm> 4 #include<cstdio> 5 #include<cmath> 6 #include<cstring> 7 #include<queue> 8 using namespace std; 9 const int INF=0x3f3f3f3f; 10 const int mxn=100010; 11 int read(){ 12 int x=0,f=1;char ch=getchar(); 13 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 14 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 15 return x*f; 16 } 17 struct edge{ 18 int v,nxt,f; 19 }e[mxn<<1]; 20 int hd[mxn],mct=1; 21 inline void add_edge(int u,int v,int f){ 22 e[++mct].v=v;e[mct].nxt=hd[u];e[mct].f=f;hd[u]=mct;return; 23 } 24 inline void insert(int u,int v,int f){ 25 // printf("ins:%d %d %d ",u,v,f); 26 add_edge(u,v,f);add_edge(v,u,0); 27 } 28 int ST,ED,S,T; 29 int n,m; 30 queue<int>q; 31 int d[mxn]; 32 bool BFS(){ 33 for(int i=0;i<=ED;i++)d[i]=0; 34 q.push(ST); 35 d[ST]=1; 36 while(!q.empty()){ 37 int u=q.front();q.pop(); 38 for(int i=hd[u];i;i=e[i].nxt){ 39 int v=e[i].v; 40 if(e[i].f && !d[v]){ 41 d[v]=d[u]+1; 42 q.push(v); 43 } 44 } 45 } 46 return d[ED]; 47 } 48 int cur[mxn]; 49 int DFS(int u,int lim){ 50 if(u==ED)return lim; 51 int f=0,tmp; 52 for(int &i=cur[u];i;i=e[i].nxt){ 53 int v=e[i].v; 54 if(e[i].f && d[v]==d[u]+1 && (tmp=DFS(v,min(lim,e[i].f)))){ 55 e[i].f-=tmp; 56 e[i^1].f+=tmp; 57 lim-=tmp; 58 f+=tmp; 59 if(!lim)return f; 60 } 61 } 62 d[u]=0; 63 return f; 64 } 65 int Dinic(){ 66 int res=0,flow=0; 67 while(BFS()){ 68 for(int i=0;i<=ED;i++)cur[i]=hd[i]; 69 while(flow=DFS(ST,INF))res+=flow; 70 } 71 return res; 72 } 73 // 74 int du[mxn]; 75 int D[mxn]; 76 int down[450][2333]; 77 bool vis[450][2333]; 78 int id[450][2333]; 79 void solve_MX(){ 80 int i,j; 81 ST=S;ED=T; 82 int ans=Dinic(); 83 for(i=1;i<=n;i++){ 84 for(j=hd[i];j;j=e[j].nxt){ 85 int v=e[j].v; 86 if(v>n && v<T){ 87 down[i][v-n]+=e[j^1].f; 88 } 89 } 90 } 91 printf("%d ",ans); 92 for(i=1;i<=n;i++){ 93 for(j=1;j<=m;j++){ 94 if(vis[i][j]) 95 printf("%d ",down[i][j]); 96 } 97 } 98 return; 99 } 100 void init(){ 101 memset(hd,0,sizeof hd);mct=1; 102 memset(vis,0,sizeof vis); 103 memset(du,0,sizeof du); 104 return; 105 } 106 int main(){ 107 freopen("in.txt","r",stdin); 108 int i,j; 109 while(scanf("%d%d",&n,&m)!=EOF){ 110 init(); 111 S=0;T=n+m+1;ST=T+1;ED=ST+1; 112 int u,v,w,L,R; 113 for(i=1;i<=m;i++){ 114 w=read(); 115 du[i+n]-=w; 116 du[T]+=w; 117 insert(i+n,T,INF-w); 118 } 119 for(i=1;i<=n;i++){ 120 int C=read();D[i]=read(); 121 insert(S,i,D[i]); 122 for(j=1;j<=C;j++){ 123 v=read()+1;L=read();R=read(); 124 du[i]-=L; 125 du[v+n]+=L; 126 down[i][v]=L; 127 vis[i][v]=1; 128 insert(i,v+n,R-L); 129 id[i][v]=mct; 130 } 131 } 132 insert(T,S,INF); 133 int smm=0; 134 for(i=S;i<=T;i++){ 135 if(du[i]>0)smm+=du[i],insert(ST,i,du[i]); 136 else insert(i,ED,-du[i]); 137 } 138 int res=Dinic();//printf("res:%d smm:%d ",res,smm); 139 if(res!=smm){ 140 puts("-1");continue; 141 } 142 else{ 143 solve_MX(); 144 } 145 } 146 return 0; 147 }