ZOJ 3229 Shoot the Bullet

Shoot the Bullet

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on  ZJU. Original ID: 3229
64-bit integer IO format: %lld      Java class name: Main

Special Judge

 

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

External Links

 
Wikipedia
Touhou Wiki

 

Source

ZOJ Monthly, July 2009

Author

WU, Zejun

 

解题：有源上下界流

 

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <cmath>

  5 #include <algorithm>

  6 #include <climits>

  7 #include <vector>

  8 #include <queue>

  9 #include <cstdlib>

 10 #include <string>

 11 #include <set>

 12 #include <stack>

 13 #define LL long long

 14 #define pii pair<int,int>

 15 #define INF 0x3f3f3f3f

 16 using namespace std;

 17 const int maxn = 1500;

 18 struct arc{

 19     int to,flow,next;

 20     arc(int x = 0,int y = 0,int z = -1){

 21         to = x;

 22         flow = y;

 23         next = z;

 24     }

 25 };

 26 int head[maxn],d[maxn],cur[maxn],tot,n,m,S,T;

 27 arc e[1000010];

 28 void add(int u,int v,int flow){

 29     e[tot] = arc(v,flow,head[u]);

 30     head[u] = tot++;

 31     e[tot] = arc(u,0,head[v]);

 32     head[v] = tot++;

 33 }

 34 bool bfs(){

 35     queue<int>q;

 36     memset(d,-1,sizeof(d));

 37     d[S] = 1;

 38     q.push(S);

 39     while(!q.empty()){

 40         int u = q.front();

 41         q.pop();

 42         for(int i = head[u]; ~i; i = e[i].next){

 43             if(e[i].flow && d[e[i].to] == -1){

 44                 d[e[i].to] = d[u] + 1;

 45                 q.push(e[i].to);

 46             }

 47         }

 48     }

 49     return d[T] > -1;

 50 }

 51 int dfs(int u,int low){

 52     if(u == T) return low;

 53     int tmp = 0,a;

 54     for(int &i = cur[u]; ~i; i = e[i].next){

 55         if(e[i].flow && d[e[i].to] == d[u] + 1 && (a=dfs(e[i].to,min(e[i].flow,low)))){

 56             tmp += a;

 57             low -= a;

 58             e[i].flow -= a;

 59             e[i^1].flow += a;

 60             if(!low) break;

 61         }

 62     }

 63     if(!tmp) d[u] = -1;

 64     return tmp;

 65 }

 66 int dinic(){

 67     int tmp = 0;

 68     while(bfs()){

 69         memcpy(cur,head,sizeof(head));

 70         tmp += dfs(S,INF);

 71     }

 72     return tmp;

 73 }

 74 int de[maxn],pos[400][maxn],low[400][maxn];

 75 int main() {

 76     int gs,day,dayup,ith,down,up;

 77     while(~scanf("%d %d",&n,&m)){

 78         memset(head,-1,sizeof(head));

 79         memset(de,0,sizeof(de));

 80         memset(low,-1,sizeof(low));

 81         int full = tot = 0;

 82         for(int i = 1; i <= m; i++){

 83             scanf("%d",&gs);

 84             add(n+i,n+m+1,INF-gs);//女孩到汇点

 85             de[n+i] -= gs;

 86             de[n+m+1] += gs;

 87         }

 88         for(int i = 1; i <= n; i++){

 89             scanf("%d %d",&day,&dayup);

 90             add(0,i,dayup);

 91             for(int j = 1; j <= day; j++){

 92                 scanf("%d %d %d",&ith,&down,&up);

 93                 pos[i][++ith] = tot;//第i天第ith个女孩

 94                 add(i,n+ith,up-down);//转化

 95                 de[i] -= down;

 96                 de[n+ith] += down;

 97                 low[i][ith] = down;

 98             }

 99         }

100         add(n+m+1,0,INF);

101         S = n+m+2;

102         T = n+m+3;

103         for(int i = 0; i <= n + m + 1; i++){

104             if(de[i] > 0){

105                 full += de[i];

106                 add(S,i,de[i]);

107             }

108             if(de[i] < 0) add(i,T,-de[i]);

109         }

110         if(dinic() == full){

111             S = 0;

112             T = n + m + 1;

113             printf("%d\n",dinic());

114             for(int i = 1; i <= n; i++){

115                 for(int j = 1; j <= m; j++)

116                     if(low[i][j] > -1) printf("%d\n",low[i][j] + e[pos[i][j]^1].flow);

117             }

118 

119         }else puts("-1");

120         puts("");

121     }

122     return 0;

123 }

View Code