您的位置:首页 > 其它

HDU 3338 Kakuro Extension

2015-08-25 13:53 316 查看
网络最大流

TLE了两天的题目。80次Submit才AC,发现是刘汝佳白书的Dinic代码还可以优化。。。。。瞬间无语。。。。。

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn = 30000 + 10;
const int INF = 0x7FFFFFFF;
struct Edge
{
int from, to, cap, flow;
Edge(int u, int v, int c, int f) :from(u), to(v), cap(c), flow(f) {}
};
vector<Edge>edges;
vector<int>G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
int n, m, s, t;

void init()
{
for (int i = 0; i < maxn; i++)
G[i].clear();
edges.clear();
}
void AddEdge(int from, int to, int cap)
{
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
int w = edges.size();
G[from].push_back(w - 2);
G[to].push_back(w - 1);
}
bool BFS()
{
memset(vis, 0, sizeof(vis));
queue<int>Q;
Q.push(s);
d[s] = 0;
vis[s] = 1;
while (!Q.empty())
{
int x = Q.front();
Q.pop();
for (int i = 0; i<G[x].size(); i++)
{
Edge e = edges[G[x][i]];
if (!vis[e.to] && e.cap>e.flow)
{
vis[e.to] = 1;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x, int a)
{
if (x == t || a == 0)
return a;
int flow = 0, f;
for (int &i = cur[x]; i<G[x].size(); i++)
{
Edge e = edges[G[x][i]];
if (d[x]+1 == d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0)
{
edges[G[x][i]].flow+=f;
edges[G[x][i] ^ 1].flow-=f;
flow+=f;
a-=f;
if(a==0) break;
}
}
if(!flow) d[x] = -1;
return flow;
}
int dinic(int s, int t)
{
int flow = 0;
while (BFS())
{
memset(cur, 0, sizeof(cur));
flow += DFS(s, INF);
}
return flow;
}

//输入输出
int N, M;
char S[1000];
int Map1[105][105];//横向
int Map2[105][105];//纵向
int BH[105][105];
int FLAG[maxn];

int main()
{
while (~scanf("%d%d", &N, &M))
{

edges.clear();
for (int i = 0; i<maxn; i++) G[i].clear();
memset(Map1, 0, sizeof(Map1));
memset(Map2, 0, sizeof(Map2));
memset(BH, 0, sizeof(BH));
for (int i = 1; i <= N; i++)
for (int j = 1; j <= M; j++)
{
scanf("%s", S);
if (S[3] == '.') continue;
if (S[3] == 'X') Map1[i][j] = -1, Map2[i][j] = -1;
else
{
if (S[2] == 'X')
{
Map2[i][j] = -1;
Map1[i][j] = (S[4] - '0') * 100 + (S[5] - '0') * 10 + (S[6] - '0') * 1;
}
else if (S[4] == 'X')
{
Map1[i][j] = -1;
Map2[i][j] = (S[0] - '0') * 100 + (S[1] - '0') * 10 + (S[2] - '0') * 1;
}
else
{
Map1[i][j] = (S[4] - '0') * 100 + (S[5] - '0') * 10 + (S[6] - '0') * 1;
Map2[i][j] = (S[0] - '0') * 100 + (S[1] - '0') * 10 + (S[2] - '0') * 1;
}
}
}
int Tot = 1;
for (int i = 1; i <= N; i++)
for (int j = 1; j <= M; j++)
if (Map1[i][j] == 0)
BH[i][j] = Tot, Tot++;
int Duan = Tot - 1;//方格编号1--Duan
s = 0;
t = 30000;
for (int i = 1; i <= N; i++)
for (int j = 1; j <= M; j++)
{
if (Map1[i][j] != -1 && Map1[i][j] != 0)
{
int Zong = 0;
for (int k = j + 1;; k++)
{
if (k>M || Map1[i][k] != 0) break;
AddEdge(Tot, BH[i][k], 8);
Zong++;
}
AddEdge(s, Tot, Map1[i][j] - Zong);
Tot++;
}
}
for (int i = 1; i <= N; i++)
for (int j = 1; j <= M; j++)
{
if (Map2[i][j] != -1 && Map2[i][j] != 0)
{
int Zong = 0;
for (int k = i + 1;; k++)
{
if (Map2[k][j] != 0 || k>N) break;
AddEdge(BH[k][j], Tot, 8);
Zong++;
}
AddEdge(Tot, t, Map2[i][j] - Zong);
Tot++;
}
}
dinic(s, t);
memset(FLAG, 0, sizeof FLAG);
for (int i = 0; i<edges.size(); i = i + 2)
if (edges[i].from >= 1 && edges[i].from <= Duan)
FLAG[edges[i].from] = edges[i].flow;
for (int i = 1; i <= N; i++)
{
for (int j = 1; j <= M; j++)
{
if (BH[i][j] == 0) printf("_");
else printf("%d", FLAG[BH[i][j]] + 1);
if (j<M) printf(" ");
}
printf("\n");
}
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: