HDU 2686 Matrix && HDU 3376 Matrix Again（最大费用）

题目地址

题意：告诉你一个矩阵，每个点有它所代表的值，每个点都只能走一次，求从( 0,0)->(n-1,n-1)->(0,0)的最大值。（两题的写法完全一样，只是第二题数据范围加大了）

思路：其实就可以转化为两次从( 0,0)->(n-1,n-1)的最大费用，就是把除了源点和汇点( 0,0)以及(n-1,n-1)的其他边的容量置为1，他们置为2，然后只能向右或者向下走，按这个要求建边，再这样跑一边最大费用就好了

HDU 2686

#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <iomanip>
#define N 610*610*2+2
#define M 800100
#define LL __int64
#define inf 0x3f3f3f3f
#define lson l,mid,ans<<1
#define rson mid+1,r,ans<<1|1
#define getMid (l+r)>>1
#define movel ans<<1
#define mover ans<<1|1
using namespace std;
const LL mod = 1000000007;
int head
, mapp[610][610];
4000
int n, m, cnt;
struct node {
int to;
int cap;//剩余流量
int money;//费用
int next;
}edge[M];
struct MCMF {
int len
;//spfa求出的长度
int pre
;//spfa求出最短路径的前缀
int path
;//因为a->b可能有重复的边，所以记录是哪条边
bool vis
;
void init() {
memset(head, -1, sizeof(head));
cnt = 0;
}
void add(int u, int v, int cap, int money) {
edge[cnt].to = v, edge[cnt].cap = cap, edge[cnt].next = head[u], edge[cnt].money = money, head[u] = cnt++;
edge[cnt].to = u, edge[cnt].cap = 0, edge[cnt].next = head[v], edge[cnt].money = -money, head[v] = cnt++;
}
bool spfa(int s, int t) {
memset(vis, false, sizeof(vis));
memset(pre, -1, sizeof(pre));
memset(path, -1, sizeof(path));
memset(len, -inf, sizeof(len));
queue<int> q;
q.push(s);
vis[s] = true;
len[s] = 0;
while (!q.empty()) {
int u = q.front();
q.pop();
vis[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if (len[v] < len[u] + edge[i].money&&edge[i].cap > 0) {
len[v] = len[u] + edge[i].money;
pre[v] = u;
path[v] = i;
if (!vis[v]) {
vis[v] = true;
q.push(v);
}
}
}
}
return len[t] != -inf;
}
int MinCostMaxFlow(int s, int t) {
LL sum = 0;
while (spfa(s, t)) {
int mmin = inf;
for (int i = t; i != s && i != -1; i = pre[i]) {
mmin = min(mmin, edge[path[i]].cap);
}
for (int i = t; i != s && i != -1; i = pre[i]) {
edge[path[i]].cap -= mmin;
edge[path[i] ^ 1].cap += mmin;
}
if (len[t] <= 0) {
break;
}
sum += mmin*(len[t]);
}
return sum;
}
};
int main() {
cin.sync_with_stdio(false);
MCMF mcmf;
int a, b, c, d;
while (~scanf("%d",&n)) {
mcmf.init();
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", &mapp[i][j]);
if ((i == 0 && j == 0) || (i == n - 1 && j == n - 1)) {
mcmf.add(i*n + j, i*n + j + n*n, 2, mapp[i][j]);//起点和终点可以走两次，点的费用为mapp[i][j](自身到自身)(当走完两边以后源点和汇点的容量都没有了所以就结束了)
}
else{
mcmf.add(i*n + j, i*n + j + n*n, 1, mapp[i][j]);
}
if (i < n - 1) {
mcmf.add(i*n + j + n*n, (i + 1)*n + j, 1, 0);//从上到下
}
if (j < n - 1) {
mcmf.add(i*n + j + n*n, i*n + j + 1, 1, 0);//从左到右
}
}
}
int s, t;
s = 2 * n*n + 1;//源点
t = 2 * n*n + 2;//汇点
mcmf.add(s, 0, 2, 0);//起点和终点可以走两次
mcmf.add(2 * n*n - 1, t, 2, 0);
printf("%d\n", mcmf.MinCostMaxFlow(s, t) - mapp[0][0] - mapp[n - 1][n - 1]);
}
return 0;
}

HDU 3376

#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <iomanip>
#define N 610*610*2+2
#define LL __int64
#define inf 0x3f3f3f3f
#define lson l,mid,ans<<1
#define rson mid+1,r,ans<<1|1
#define getMid (l+r)>>1
#define movel ans<<1
#define mover ans<<1|1
using namespace std;
const LL mod = 1000000007;
int head
, mapp[610][610];
int n, m, cnt;
struct node {
int to;
int cap;//剩余流量
int money;//费用
int next;
}edge[N << 2];
int len
;//spfa求出的长度
int pre
;//spfa求出最短路径的前缀
int path
;//因为a->b可能有重复的边，所以记录是哪条边
bool vis
;
struct MCMF {
void init() {
memset(head, -1, sizeof(head));
cnt = 0;
}
void add(int u, int v, int cap, int money) {
edge[cnt].to = v, edge[cnt].cap = cap, edge[cnt].next = head[u], edge[cnt].money = money, head[u] = cnt++;
edge[cnt].to = u, edge[cnt].cap = 0, edge[cnt].next = head[v], edge[cnt].money = -money, head[v] = cnt++;
}
bool spfa(int s, int t) {
memset(vis, false, sizeof(vis));
memset(pre, -1, sizeof(pre));
memset(path, -1, sizeof(path));
memset(len, -inf, sizeof(len));
queue<int> q;
q.push(s);
vis[s] = true;
len[s] = 0;
while (!q.empty()) {
int u = q.front();
q.pop();
vis[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if (len[v] < len[u] + edge[i].money&&edge[i].cap > 0) {
len[v] = len[u] + edge[i].money;
pre[v] = u;
path[v] = i;
if (!vis[v]) {
vis[v] = true;
q.push(v);
}
}
}
}
return len[t] != -inf;
}
int MinCostMaxFlow(int s, int t) {
LL sum = 0;
while (spfa(s, t)) {
int mmin = inf;
for (int i = t; i != s && i != -1; i = pre[i]) {
mmin = min(mmin, edge[path[i]].cap);
}
for (int i = t; i != s && i != -1; i = pre[i]) {
edge[path[i]].cap -= mmin;
edge[path[i] ^ 1].cap += mmin;
}
if (len[t] <= 0) {
break;
}
sum += mmin*len[t];
}
return sum;
}
};
int main() {
cin.sync_with_stdio(false);
MCMF mcmf;
int a, b, c, d;
while (~scanf("%d",&n)) {
mcmf.init();
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", &mapp[i][j]);
if ((i == 0 && j == 0) || (i == n - 1 && j == n - 1)) {
mcmf.add(i*n + j, i*n + j + n*n, 2, mapp[i][j]);//起点和终点可以走两次，点的费用为mapp[i][j](自身到自身)(当走完两边以后源点和汇点的容量都没有了所以就结束了)
}
else{
mcmf.add(i*n + j, i*n + j + n*n, 1, mapp[i][j]);
}
if (i < n - 1) {
mcmf.add(i*n + j + n*n, (i + 1)*n + j, 1, 0);//从上到下
}
if (j < n - 1) {
mcmf.add(i*n + j + n*
c551
n, i*n + j + 1, 1, 0);//从左到右
}
}
}
int s, t;
s = 2 * n*n + 1;//源点
t = 2 * n*n + 2;//汇点
mcmf.add(s, 0, 2, 0);//起点和终点可以走两次
mcmf.add(2 * n*n - 1, t, 2, 0);
printf("%d\n", mcmf.MinCostMaxFlow(s, t) - mapp[0][0] - mapp[n - 1][n - 1]);
}
return 0;
}