題干：

A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= p?max(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c?max(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= l?max(u,v) of power delivered by u to v. Let Con=Σ?uc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.?

An example is in figure 1. The label x/y of power station u shows that p(u)=x and pmax(u)=y. The label x/y of consumer u shows that c(u)=x and c?max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l?max(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.?

Input

There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of l?max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p?max(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of c?max(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.

Output

For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.

Sample Input

2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20 7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7(3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5(0)5 (1)2 (3)2 (4)1 (5)4

Sample Output

15 6

Hint

The sample input contains two data sets. The first data set encodes a network with 2 nodes, power station 0 with pmax(0)=15 and consumer 1 with cmax(1)=20, and 2 power transport lines with lmax(0,1)=20 and lmax(1,0)=10. The maximum value of Con is 15. The second data set encodes the network from figure 1.

題目大意：

簡(jiǎn)單的說(shuō)下題意（按輸入輸出來(lái)講，前面的描述一堆的rubbish，還用來(lái)誤導(dǎo)人），給你n個(gè)點(diǎn)，其中有np個(gè)是能提供電力的點(diǎn)，nc個(gè)是能消費(fèi)電力的點(diǎn)，剩下的點(diǎn)(n-np-nc)是中轉(zhuǎn)戰(zhàn)即不提供電力也不消費(fèi)電力，點(diǎn)與點(diǎn)之間是有線路存在的，有m條線路，每條線路有最多運(yùn)載限定。?
前4個(gè)數(shù)據(jù)就是有n個(gè)點(diǎn)，np個(gè)供電點(diǎn),nc個(gè)消費(fèi)點(diǎn),m條線路，接來(lái)題目先給出的是m條線路的數(shù)據(jù)，(起點(diǎn),終點(diǎn))最多運(yùn)載量,然后是np個(gè)供電點(diǎn)的數(shù)據(jù)(供電點(diǎn))最多供電量，接著就是nc個(gè)消費(fèi)點(diǎn)的數(shù)據(jù)(消費(fèi)點(diǎn))最多消費(fèi)電量。?
題目要我們求出給定的圖最大能消費(fèi)的總電量（就是求最大流）

解題報(bào)告：

? 模板題。

AC代碼：

#include<cstring> #include<cstdio> #include<algorithm> #include<iostream> using namespace std; int tot; struct Edge {int to,ne,w; } e[100005 * 2]; int head[10005]; int st,ed; int dis[10050],q[10005];//一共多少個(gè)點(diǎn)跑bfs，dis數(shù)組和q數(shù)組就開(kāi)多大。 void add(int u,int v,int w) {e[++tot].to=v; e[tot].w=w; e[tot].ne=head[u]; head[u]=tot;e[++tot].to=u; e[tot].w=0; e[tot].ne=head[v]; head[v]=tot; } bool bfs(int st,int ed) {memset(dis,-1,sizeof(dis));int front=0,tail=0;q[tail++]=st;dis[st]=0;while(front<tail) {int cur = q[front];if(cur == ed) return 1;front++;for(int i = head[cur]; i!=-1; i = e[i].ne) {if(e[i].w&&dis[e[i].to]<0) {q[tail++]=e[i].to;dis[e[i].to]=dis[cur]+1;}}}if(dis[ed]==-1) return 0;return 1; } int dfs(int cur,int limit) {//limit為源點(diǎn)到這個(gè)點(diǎn)的路徑上的最小邊權(quán) if(limit==0||cur==ed) return limit;int w,flow=0;for(int i = head[cur]; i!=-1; i = e[i].ne) { if(e[i].w&&dis[e[i].to]==dis[cur]+1) {w=dfs(e[i].to,min(limit,e[i].w));e[i].w-=w;e[i^1].w+=w;flow+=w;limit-=w;if(limit==0) break;}}if(!flow) dis[cur]=-1;return flow; } int dinic() {int ans = 0;while(bfs(st,ed)) ans+=dfs(st,0x7fffffff);return ans; } inline int read() {char ch = getchar(); int x = 0, f = 1;while(ch < '0' || ch > '9') {if(ch == '-') f = -1;ch = getchar();} while('0' <= ch && ch <= '9') {x = x * 10 + ch - '0';ch = getchar();} return x * f; } int main() {int n,np,nc,m;while(~scanf("%d%d%d%d",&n,&np,&nc,&m)) {tot=1;for(int i = 0; i<=n+2; i++) head[i] = -1;st=n+1,ed=n+2; for(int u,v,w,i = 1; i<=m; i++) {u=read();v=read();w=read();add(u,v,w);}for(int u,w,i = 1; i<=np; i++) {u=read();w=read();add(st,u,w);}for(int u,w,i = 1; i<=nc; i++) {u=read();w=read();add(u,ed,w);}printf("%d\n",dinic()); }return 0; }

?

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