poj 3422 Kaka‘s Matrix Travels — K取方格数（最大费用最大流）

Posted 2021-10-07 jpphy0

目录

• 问题
• 分析
• 代码

问题

K取方格数 - http://poj.org/problem?id=3422

分析

代码

/* K取方格 poj3422 */
#include<bits/stdc++.h>
using namespace std;
const int mxv = 210, mxe = 210<<2, inf = 0x3f3f3f3f, mxn = 10;
int n, m, k, ver[mxn][mxn][2];
int pre[mxv], head[mxv], tot, d[mxe], flow[mxv], v[mxe];
struct Edge{ int to, nxt, cap, w;}edge[mxe];
void addEdge(int u, int v, int cap, int w){
	Edge &e = edge[++tot];
	e.nxt = head[u], e.to = v, e.cap = cap, e.w = w, head[u] = tot;
}
void add(int u, int v, int cap, int w){
	addEdge(u, v, cap, w), addEdge(v, u, 0, -w);
}
bool spfa(int s, int t){
    queue<int> q;
	int f;
	memset(d, 0xcf, sizeof d), memset(v, 0, sizeof v);
    v[s] = 1, d[s] = 0, flow[s] = inf, q.push(s);
    while(!q.empty()){
        f = q.front(), q.pop(), v[f] = 0;
        for(int to, i = head[f]; ~i; i = edge[i].nxt){
			if(!edge[i].cap) continue;
            to = edge[i].to;
            if(d[to] < d[f] + edge[i].w){
                d[to] = d[f] + edge[i].w;
				flow[to] = min(flow[f], edge[i].cap);
				pre[to] = i;
                if(!v[to]) v[to] = 1, q.push(to);
            }
        }
    }
	return d[t] != 0xcfcfcfcf;
}
void update(int s, int t){
	for(int i, now = t; now != s; now = edge[i^1].to)
		i = pre[now], edge[i].cap -= flow[t], edge[i^1].cap += flow[t];
}
int main(){
	int t = 0, S = 201, T = S+1, w, ans;
	for(int i = 0; i < mxn; ++i)
		for(int j = 0; j < mxn; ++j)
			for(int k = 0; k <= 1; ++k) ver[i][j][k] = ++t;
	scanf("%d", &t);
	while(t--){
		tot = 1, ans = 0, memset(head, -1, sizeof head);
		scanf("%d%d%d", &n, &m, &k);
		for(int i = 0; i < n; ++i){
			for(int j = 0; j < m; ++j){
				scanf("%d", &w);
				add(ver[i][j][0], ver[i][j][1], 1, w), add(ver[i][j][0], ver[i][j][1], k, 0);
				if(j) add(ver[i][j-1][1], ver[i][j][0], k, 0);
				if(i) add(ver[i-1][j][1], ver[i][j][0], k, 0);
			}			
		}
		add(S, ver[0][0][0], k, 0), add(ver[n-1][m-1][1], T, k, 0);
		while(spfa(S, T)) ans += flow[T]*d[T], update(S, T);
		printf("%d\\n", ans);
	}	
    return 0;
}