比较简单的题了。
只需从左上角到右下角找两条路就可以了。
因为每个点只能走一次，所以拆点，限制流量为1。
因为求的是最大值，所以权值取反求最小值。
因为第一个点和最后一个点经过两次，只算一次，最后要减去。
ps：数组还是开大点好。。。。不知道什么时候就SB了。。。
注意汇点可能不是最后一个点（模板的问题。。
注意初始化的范围。。。。
matrix again只是数据大了，算法不用改。。。无聊。。。。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <bitset>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <list>
#include <map>
#include <set>
#define pk(x) printf("%d\n", x)
using namespace std;
#define PI acos(-1.0)
#define EPS 1E-6
#define clr(x,c) memset(x,c,sizeof(x))
//#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
const int MAXV = 2000;
const int INF = 1<<30;
struct Edge { int to, cap, cost, rev; };
vector<Edge> G[MAXV];
int dist[MAXV], prv[MAXV], pre[MAXV], in[MAXV];
queue<int> que;
void addedge(int from, int to, int cap, int cost) {
G[from].push_back((Edge){to, cap, cost, G[to].size()});
G[to].push_back((Edge){from, 0, -cost, G[from].size()-1});
}
int min_cost_max_flow(int s, int t) { //, int f) {
int res = 0;
int f = 0;
while (1) { //f > 0) {
for (int i = 1; i <= t; ++i) dist[i] = INF, in[i] = 0;
dist[s] = 0;
while (!que.empty()) que.pop();
in[s] = 1;
que.push(s);
while (!que.empty()) {
int u = que.front(); que.pop(); in[u] = 0;
for (int i = 0; i < G[u].size(); ++i) {
Edge &e = G[u][i];
if (e.cap > 0 && dist[e.to] > dist[u] + e.cost) {
dist[e.to] = dist[u] + e.cost;
prv[e.to] = u;
pre[e.to] = i;
if (in[e.to] == 0) {
in[e.to] = 1;
que.push(e.to);
}
}
}
}
if (dist[t] == INF) break; //return -1;
int d = INF; // d = f;
for (int v = t; v != s; v = prv[v]) {
d = min(d, G[prv[v]][pre[v]].cap);
}
f += d;
res += d * dist[t];
for (int v = t; v != s; v = prv[v]) {
Edge &e = G[prv[v]][pre[v]];
e.cap -= d;
G[v][e.rev].cap += d;
}
}
return res;
//return f;
}
int n;
int getid1(int x, int y) { return x*n+y+1; }
int getid2(int x, int y) { return x*n+y+n*n+1; }
int a[30][30];
int main()
{
while (~scanf("%d", &n)) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
scanf("%d", &a[i][j]);
}
}
for (int i = 0; i <= 2*n*n; ++i) G[i].clear();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i+j == 0) addedge(getid1(i,j), getid2(i,j), 2, -a[i][j]);
else if (i+j == n*2 - 2) addedge(getid1(i,j), getid2(i, j), 2, -a[i][j]);
else addedge(getid1(i, j), getid2(i, j), 1, -a[i][j]);
if (i+1<n) addedge(getid2(i, j), getid1(i+1, j), 1, 0);
if (j+1<n) addedge(getid2(i, j), getid1(i, j+1), 1, 0);
}
}
printf("%d\n", -min_cost_max_flow(getid1(0,0), getid2(n-1,n-1)) - a[0][0] - a[n-1][n-1]);
}
return 0;
}
hdu2686
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <bitset>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <list>
#include <map>
#include <set>
#define pk(x) printf("%d\n", x)
using namespace std;
#define PI acos(-1.0)
#define EPS 1E-6
#define clr(x,c) memset(x,c,sizeof(x))
//#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
const int MAXV = 610*610*2+2;
const int INF = 1<<30;
struct Edge { int to, cap, cost, rev; };
vector<Edge> G[MAXV];
int dist[MAXV], prv[MAXV], pre[MAXV], in[MAXV];
queue<int> que;
void addedge(int from, int to, int cap, int cost) {
G[from].push_back((Edge){to, cap, cost, G[to].size()});
G[to].push_back((Edge){from, 0, -cost, G[from].size()-1});
}
int min_cost_max_flow(int s, int t) { //, int f) {
int res = 0;
int f = 0;
while (1) { //f > 0) {
for (int i = 1; i <= t; ++i) dist[i] = INF, in[i] = 0;
dist[s] = 0;
while (!que.empty()) que.pop();
in[s] = 1;
que.push(s);
while (!que.empty()) {
int u = que.front(); que.pop(); in[u] = 0;
for (int i = 0; i < G[u].size(); ++i) {
Edge &e = G[u][i];
if (e.cap > 0 && dist[e.to] > dist[u] + e.cost) {
dist[e.to] = dist[u] + e.cost;
prv[e.to] = u;
pre[e.to] = i;
if (in[e.to] == 0) {
in[e.to] = 1;
que.push(e.to);
}
}
}
}
if (dist[t] == INF) break; //return -1;
int d = INF; // d = f;
for (int v = t; v != s; v = prv[v]) {
d = min(d, G[prv[v]][pre[v]].cap);
}
f += d;
res += d * dist[t];
for (int v = t; v != s; v = prv[v]) {
Edge &e = G[prv[v]][pre[v]];
e.cap -= d;
G[v][e.rev].cap += d;
}
}
return res;
//return f;
}
inline int Scan()
{
char ch = getchar();
int data = 0;
while (ch < ‘0‘ || ch > ‘9‘) ch = getchar();
do {
data = data*10 + ch-‘0‘;
ch = getchar();
} while (ch >= ‘0‘ && ch <= ‘9‘);
return data;
}
int n;
int getid1(int x, int y) { return x*n+y+1; }
int getid2(int x, int y) { return x*n+y+n*n+1; }
int a[600][600];
int main()
{
while (~scanf("%d", &n)) {
int maxn = 2*n*n;
for (int i = 0; i <= maxn; ++i) G[i].clear();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
a[i][j] = Scan();
if (i+j == 0 || i+j == n*2-2) addedge(getid1(i,j), getid2(i,j), 2, -a[i][j]);
else addedge(getid1(i, j), getid2(i, j), 1, -a[i][j]);
if (i+1<n) addedge(getid2(i, j), getid1(i+1, j), 1, 0);
if (j+1<n) addedge(getid2(i, j), getid1(i, j+1), 1, 0);
}
}
printf("%d\n", -min_cost_max_flow(getid1(0,0), getid2(n-1,n-1)) - a[0][0] - a[n-1][n-1]);
}
return 0;
}
hdu3376
时间： 2024-12-14 13:48:41