[codevs 1917] 深海机器人问题

[codevs 1917] 深海机器人问题

题解：

看题建图。

“k个深海机器人从(x,y)位置坐标出发”、“r个深海机器人可选择(x,y)位置坐标作为目的地”，嗯~暗示已经很清楚了。

一开始没理解题目中关于地图大小的叙述，还要记得建立多重边（有点方格取数2的味道）。

代码：

总时间耗费: 5ms 
总内存耗费: 364B

#include<cstdio>
#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn = 500 + 10;
const int INF = 1e9 + 7;

struct Edge {
	int from, to, cap, flow, cost;
};

vector<Edge> edges;
vector<int> G[maxn];

void AddEdge(int from, int to, int cap, int cost) {
	edges.push_back((Edge){from, to, cap, 0, cost});
	edges.push_back((Edge){to, from, 0, 0, -cost});
	int m = edges.size();
	G[from].push_back(m-2);
	G[to].push_back(m-1);
}

int s, t;
int ID[maxn][maxn];

int d[maxn*maxn], p[maxn*maxn], a[maxn*maxn];
bool inq[maxn];

bool BellmanFord(int& cost) {
	for(int i = s; i <= t; i++) d[i] = INF;
	memset(inq, 0, sizeof(inq));
	d[s] = 0; inq[s] = 1; a[s] = INF; p[s] = 0;
	
	queue<int> Q;
	Q.push(s);
	while(!Q.empty()) {
		int x = Q.front(); Q.pop();
		inq[x] = 0;
		for(int i = 0; i < G[x].size(); i++) {
			Edge& e = edges[G[x][i]];
			if(e.cap > e.flow && d[e.to] > d[x] + e.cost) {
				d[e.to] = d[x] + e.cost;
				p[e.to] = G[x][i];
				a[e.to] = min(a[x], e.cap-e.flow);
				if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
			}
		}
	}
	if(d[t] == INF) return 0;
	
	cost += a[t] * d[t];
	
	int x = t;
	while(x != s) {
		edges[p[x]].flow += a[t];
		edges[p[x]^1].flow -= a[t];
		x = edges[p[x]].from;
	}
	
	return 1;
}

void MincostMaxflow() {
	int cost = 0;
	while(BellmanFord(cost));
	cout << -cost << endl;
}

int main() {
	int A, B, m, n;
	
	cin >> A >> B >> m >> n; 
	m++; n++; s = 0; t = m*n + 1;
	
	for(int x = 1, c = 1; x <= m; x++)
		for(int y = 1; y <= n; y++, c++)
			ID[x][y] = c;
	
	for(int x = 1; x <= m; x++)
		for(int y = 1; y < n; y++) {
			int& from = ID[x][y];
			int& to = ID[x][y+1];
			int cost;
			cin >> cost;
			AddEdge(from, to, 1, -cost);
			AddEdge(from, to, INF, 0);
		}
		
	for(int y = 1; y <= n; y++)
		for(int x = 1; x < m; x++) {
			int& from = ID[x][y];
			int& to = ID[x+1][y];
			int cost;
			cin >> cost;
			AddEdge(from, to, 1, -cost);
			AddEdge(from, to, INF, 0);
		}
	
	for(int i = 1; i <= A; i++) {
		int k, x, y;
		cin >> k >> x >> y;
		AddEdge(s, ID[x+1][y+1], k, 0);
	}
	
	for(int i = 1; i <= B; i++) {
		int r, x, y;
		cin >> r >> x >> y;
		AddEdge(ID[x+1][y+1], t, r, 0);
	}
	
	MincostMaxflow();
	
	return 0;
}