POJ

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 p max(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.

思路：这是一道最大流的模板题，用Dinic算法来写的话100ms左右就行了，直接建图，s和发电的连起来，用户和 t 连起来，容量分别为发电的量和用户需要的量，再把有连接的两个点连起来，就行了，直接套模板。  ！！！还有要注意的一点，输入的时候记得前面加个空格！！！

代码：

#include <cstdio>#include <cmath>#include <iostream>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <map>#include <numeric>#include <set>#include <string>#include <cctype>#include <sstream>#define INF 0x3f3f3f3f#define lson l, m, rt<<1#define rson m+1, r, rt<<1|1typedef long long LL;using namespace std;const int maxn = 1e5 + 5;int n,np,nc,m,s,t;struct edge{int  to , cap , rev ; };vector<edge> G[maxn];int level[maxn];int iter[maxn];void addedge(int from,int to,int cap){    G[from].push_back( (edge) { to,cap,G[to].size() } );    G[to].push_back( (edge){from,0,G[from].size()-1 } );}void bfs(int s){    memset(level,-1,sizeof(level));    queue<int>que;    level[s]=0;    que.push(s);    while (!que.empty()){        int v=que.front();que.pop();        for (int i=0;i<G[v].size();i++){            edge &e=G[v][i];            if (e.cap>0 && level[e.to]<0){                level[e.to]=level[v]+1;                que.push(e.to);            }        }    }}int dfs(int v,int t,int f){    if (v==t) return f;    for (int &i=iter[v];i<G[v].size();i++){        edge &e=G[v][i];        if (e.cap>0&&level[v]<level[e.to]){            int d=dfs(e.to,t,min(f,e.cap));            if (d>0){                e.cap-=d;                G[e.to][e.rev].cap+=d;                return d;            }        }    }    return 0;}int Maxflow(int s,int t){    int flow=0;    for (;;){        bfs(s);        if (level[t]<0) return flow;        memset(iter, 0 , sizeof(iter));        int f;        while ( (f=dfs( s , t , INF ) )>0){            flow+=f;        }    }}int main (){    //freopen ("in.txt","r",stdin);    int u,v,z;    while(~scanf("%d%d%d%d", &n, &np, &nc, &m)){        for (int i=0;i<=t;i++) G[i].clear();        s = n ; t = s + 1;        for(int i = 0; i < m; ++i){            scanf(" (%d,%d)%d", &u, &v, &z);            addedge(u, v, z);        }        for(int i = 0; i < np; ++i){            scanf(" (%d)%d", &u, &z);            addedge(s, u, z);        }        for(int i = 0; i < nc; ++i){            scanf(" (%d)%d", &u, &z);            addedge(u, t, z);        }        printf("%d\n", Maxflow(s,t));    }    return 0;}