POJ 2175 Evacuation Plan 费用流 负圈定理

题目给了一个满足最大流的残量网络，判断是否费用最小。

如果残量网络中存在费用负圈，那么不是最优，在这个圈上增广，增广1的流量就行了。

1.SPFA中某个点入队超过n次，说明存在负环，但是这个点不一定在负环上。

2.这个负环可能包括汇点t,所以构建残量网络的时候也要考虑防空洞到t上的容量。

 

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("in.txt","r",stdin);
   // freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}

const int maxn = 110;
int X[maxn], Y[maxn], B[maxn];
int P[maxn], Q[maxn], C[maxn];
int E[maxn][maxn];
int N, M;

bool input()
{
    if(scanf("%d%d", &N, &M) == EOF) return false;
    for(int i = 1; i <= N; i++)
        scanf("%d%d%d", &X[i], &Y[i], &B[i]);
    for(int i = 1; i <= M; i++)
        scanf("%d%d%d", &P[i], &Q[i], &C[i]);
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= M; j++)
            scanf("%d", &E[i][j]);
    return true;
}

struct Edge
{
    int from, to, cost, cap;
};
const int maxv = maxn * 2;
int n,s,t;
vector<int> g[maxv];
vector<Edge> edge;
int road[maxv];
int d[maxv];
int inq[maxv];
int vcnt[maxv];

int SPFA()
{
    queue<int> q;
    memset(d, INF, sizeof(d));
    memset(inq, false, sizeof(inq));
    memset(vcnt, 0, sizeof(vcnt));
    d[s] = 0;
    road[s] = -1;
    q.push(s);

    while(!q.empty())
    {
        int u = q.front(); q.pop();
        inq[u] = false;
        for(int i = 0; i < g[u].size(); i++)
        {
            Edge &e = edge[g[u][i]];
            if(e.cap > 0 && d[e.to] > d[u] + e.cost)
            {
                d[e.to] = d[u] + e.cost;
                road[e.to] = g[u][i];
                if(!inq[e.to])
                {
                    inq[e.to] = true;
                    if(++vcnt[e.to] > n) return e.to;
                    q.push(e.to);
                }
            }
        }
    }
    return -1;
}
void add(int from, int to, int cost, int cap, int flow)
{
    edge.push_back((Edge){from, to, cost, cap - flow});
    g[from].push_back(edge.size() - 1);
    edge.push_back((Edge){to, from, -cost, flow});
    g[to].push_back(edge.size() - 1);
}

int Cost(int i, int j)
{
    return abs(X[i] - P[j]) + abs(Y[i] - Q[j]) + 1;
}

void init()
{
    for(int i = 1; i <= n; i++)
        g[i].clear();
    edge.clear();
}
void construct()
{
    n = N + M + 2;
    s = n - 1;
    t = n;
    init();

    int sum[maxn] = {0};
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= M; j++)
            sum[j] += E[i][j];
    for(int i = 1; i <= N; i++)
        add(s, i, 0, B[i], 0);
    for(int j = 1; j <= M; j++)
        add(j + N, t, 0, C[j], sum[j]);
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= M; j++)
            add(i, j + N, Cost(i, j), min(B[i], C[i]), E[i][j]);
}

int vis[maxv];
void solve()
{
    int u = SPFA();
    if(u == -1)
    {
        printf("OPTIMAL\n");
        return ;
    }

    memset(vis, 0, sizeof(vis));
    while(1)
    {
        if(vis[u]) break;
        vis[u] = true;
        u = edge[road[u]].from;
    }

    memset(vis, 0, sizeof(vis));
    for(int e = road[u]; !vis[edge[e].to]; e = road[edge[e].from])
    {
        int x = edge[e].from;
        int y = edge[e].to;
        vis[y] = true;
        if(x == t || y == t) continue;
        if(x < y) E[x][y - N] += 1;
        else E[y][x - N] -= 1;
    }

    printf("SUBOPTIMAL\n");
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= M; j++)
            printf("%d%c", E[i][j], j == M? '\n': ' ');
}
int main()
{
    debug();
    while(input())
    {
        construct();
        solve();
    }
    return 0;
}