poj 1459 Power Network（网络流_最大流）

题目链接：http://poj.org/PRoblem?id=1459

Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 27853   Accepted: 14481 Description 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) <= pmax(u) of power, may consume an amount 0 <= c(u) <= min(s(u),cmax(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) <= lmax(u,v) of power delivered by u to v. Let Con=Σuc(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 cmax(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and lmax(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 lmax(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of pmax(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 cmax(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)207 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 156 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. Source Southeastern Europe 2003

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题意：给出发电站，和用户，还有各线路能流的流量，求能传给用户最大的电量

解析：就是建一个图，0~n-1为节点，n为源点，n+1为汇点

代码：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<queue>#include<cmath>#include<map>#include<vector>#include<stack>#define N 209using namespace std;const int INF = 0x3f3f3f3f;int mp[N][N], n;int max_flow(){    int ans = 0, used[N], pre[N];    n = n + 2;    while(1)    {        memset(used, 0, sizeof(used));        memset(pre, -1, sizeof(pre));        used[n - 2] = 1;        queue<int> q;        q.push(n - 2);        while(!q.empty())        {            int u = q.front(); q.pop();            if(u == n - 1) break;            for(int i = 0; i < n; i++)            {                if(!used[i] && mp[u][i])                {                    used[i] = 1;                    pre[i] = u;                    q.push(i);                }            }        }        if(pre[n - 1] == -1) break;        int m = INF;        for(int i = n - 1; pre[i] != -1; i = pre[i])        {            m = min(m, mp[pre[i]][i]);        }        ans += m;        for(int i = n - 1; pre[i] != -1;  i = pre[i])        {            mp[pre[i]][i] -= m;            mp[i][pre[i]] += m;        }    }    return ans;}int main(){    int np, nc, m, u, v, w;    while(~scanf("%d%d%d%d", &n, &np, &nc, &m))    {        memset(mp, 0, sizeof(mp));        for(int i = 0; i < m; i++)        {            scanf(" (%d,%d)%d", &u, &v, &w);            mp[u][v] += w;        }        u = n;        for(int i = 0; i < np; i++)        {            scanf(" (%d)%d", &v, &w);            mp[u][v] += w;        }        v = n + 1;        for(int i = 0; i < nc; i++)        {            scanf(" (%d)%d", &u, &w);            mp[u][v] += w;        }        printf("%d/n", max_flow());    }    return 0;}