深海机器人问题(费用流)

最新推荐文章于 2024-03-05 21:59:37 发布

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最新推荐文章于 2024-03-05 21:59:37 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/IMGavin/p/6411799.html

//http://www.cnblogs.com/IMGavin/
#include <iostream>
#include <stdio.h>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <bitset>
#include <algorithm>
using namespace std;

typedef long long LL;
#define gets(A) fgets(A, 1e8, stdin)
const int INF = 0x3F3F3F3F, N = 1000, MOD = 1003, M = 1000000;

const double EPS = 1e-6;


int ri[N][N], up[N][N];

struct Node{
	int u, v, cap, cost;
	int next;
}edge[M];//有向图，u到v的容量，费用
int tot;
int head[N], pre[N], path[N], dis[N];
bool inq[N];

void init(){
	tot = 0;
	memset(head, -1, sizeof(head));
}

void add(int u, int v, int cap, int cost){
	edge[tot].u = u;
	edge[tot].v = v;
	edge[tot].cap = cap;
	edge[tot].cost = cost;
	edge[tot].next = head[u];
	head[u] = tot++;
	edge[tot].u = v;
	edge[tot].v = u;
	edge[tot].cap = 0;
	edge[tot].cost = -cost;
	edge[tot].next = head[v];
	head[v] = tot++;
}

bool SPFA(int st, int des){//计算最小费用
	memset(inq, 0, sizeof(inq));
	memset(dis, 0x3f, sizeof(dis));
	queue <int> q;
	q.push(st);
	dis[st] = 0;
	inq[st] = true;
	while(!q.empty()){
		int u = q.front();
		q.pop();
		inq[u] = false;
		for(int i = head[u]; ~i; i = edge[i].next){
			int v = edge[i].v;
			if(edge[i].cap > 0 && dis[v] > dis[u] + edge[i].cost){
				dis[v] = dis[u] + edge[i].cost;
				pre[v] = u;
				path[v] = i;
				if(!inq[v]){
					inq[v] = true;
					q.push(v);
				}
			}
		}
	}
	return dis[des] < INF;
}

int EdmondsKarp(int st, int des){//最小费用最大流
	int mincost = 0, flow = 0;//最小费用与流量
	while(SPFA(st, des)){
		int f = INF;
		for(int i = des; i != st; i = pre[i]){
            if(f > edge[path[i]].cap){
                f = edge[path[i]].cap;
            }
		}
		for(int i = des; i != st; i = pre[i]){
			edge[path[i]].cap -= f;
			edge[path[i]^1].cap += f;
		}
		mincost += f * dis[des];
		flow += f;
	}
	return mincost;
}

int main(){
	int a, b;
	int p, q;
	while(cin >> a >> b){
		init();
		cin >> p >> q;
		for(int i = 0; i <= p; i++){
			for(int j = 0; j < q; j++){
				scanf("%d", &ri[i][j]);
			}
		}
		for(int i = 0; i <= q; i++){
			for(int j = 0; j < p; j++){
				scanf("%d", &up[j][i]);
			}
		}
		p++;
		q++;
		int st = p * q, des = st + 1;
		for(int i = 0; i < a; i++){
			int k, x, y;
			cin >> k >> x >> y;
			add(st, x * q + y, k, 0);
		}
		for(int i = 0; i < b; i++){
			int k, x, y;
			cin >> k >> x >> y;
			add(x * q + y, des, k, 0);
		}
		for(int i = 0; i <= p - 1; i++){
			for(int j = 0; j <= q - 1; j++){

				if(i < p - 1){
					add(i * q + j, (i + 1) * q + j, 1, -up[i][j]);
					add(i * q + j, (i + 1) * q + j, INF, 0);
				}

				if(j < q - 1){
					add(i * q + j, i * q + j + 1, 1, -ri[i][j]);
					add(i * q + j, i * q + j + 1, INF, 0);
				}
				
			}
		}
		printf("%d\n", -EdmondsKarp(st, des));
	}

	return 0;
}

转载于:https://www.cnblogs.com/IMGavin/p/6411799.html