这是一道变形题，不得不说是好题啊
题目大意是，有一张n*m的地图，每个点上可能是数字，代表矿石的数目，可能是*，表示一个传送阵，送往某个坐标，可能是#，代表不通。每次矿车只能往右方或者下方走一格，那么这就可以转化为一个有向图了。
每个点，往其右方和下方相邻的点建有向边，如果是#，就不建边了，如果是*，就要把*的位置跟其传送的位置建一条边。
之后就要求强连通分量，然后缩点了，再之后就是DP求最优解了，跟3160是一致的，按照拓扑序逆序DP，最后输出的是起点所在的新点的DP值
2012年7月1日：重写了此题，用topsort无数个wa,只好用spfa最长路水过,然后终于用topsort水过
依然是用Kosaraju
/*
ID: sdj22251
PROG: subset
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define LOCA
#define MAXN 5005
#define INF 100000000
#define eps 1e-7
using namespace std;
struct Edge
{
int v, next;
}edge[10 * MAXN], revedge[10 * MAXN], newedge[MAXN];
int head[MAXN], revhead[MAXN], e, visited[MAXN], newhead[MAXN];
int order[MAXN], cnt, id[MAXN], val[MAXN], newval[MAXN];
int uu[5 * MAXN], vv[5 * MAXN], out[MAXN], dp[MAXN], num;
int n, m;
int ans;
char mp[50][50];
struct Point
{
int x, y;
}p[MAXN];
void init()
{
e = 0;
cnt = 0;
num = 0;
memset(head, -1, sizeof(head));
memset(revhead, -1, sizeof(revhead));
memset(newhead, -1, sizeof(newhead));
memset(out, 0 , sizeof(out));
memset(newval, 0, sizeof(newval));
memset(dp, 0, sizeof(dp));
memset(val, 0, sizeof(val));
}
void insert(const int &x, const int &y)
{
edge[e].v = y;
edge[e].next = head[x];
head[x] = e;
revedge[e].v = x;
revedge[e].next = revhead[y];
revhead[y] = e;
e++;
}
void newinsert(const int &x, const int &y)
{
newedge[e].v = y;
newedge[e].next = newhead[x];
newhead[x] = e++;
}
void readdata()
{
for(int i = 0; i < n; i++)
scanf("%s", mp[i]);
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
if(mp[i][j] == '#') continue;
if(mp[i][j] == '*')
{
p[cnt].x = i;
p[cnt++].y = j;
}
else
{
val[i * m + j + 1] = mp[i][j] - '0';
}
if(j < m - 1)
{
if(mp[i][j + 1] != '#')
{
uu[num] = i * m + j + 1;
vv[num] = i * m + j + 2;
insert(uu[num], vv[num]);
num++;
}
}
if(i < n - 1)
{
if(mp[i + 1][j] != '#')
{
uu[num] = i * m + j + 1;
vv[num] = (i + 1) * m + j + 1;
insert(uu[num], vv[num]);
num++;
}
}
}
}
for(int i = 0; i < cnt; i++)
{
int x, y;
scanf("%d%d", &x, &y);
int xx = p[i].x;
int yy = p[i].y;
uu[num] = xx * m + yy + 1;
vv[num] = x * m + y + 1;
if(mp[x][y] != '#')
{
insert(uu[num], vv[num]);
num++;
}
}
}
void dfs(int u)
{
visited[u] = 1;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].v;
if(!visited[v])
dfs(v);
}
order[cnt++] = u;
}
void dfs_rev(int u)
{
visited[u] = 1;
id[u] = cnt;
if(val[u] > 0) newval[cnt] += val[u];
for(int i = revhead[u]; i != -1; i = revedge[i].next)
{
int v = revedge[i].v;
if(!visited[v])
dfs_rev(v);
}
}
void Kosaraju()
{
init();
readdata();
memset(visited, 0, sizeof(visited));
cnt = 0;
for(int i = 1; i <= n * m; i++)
{
if(!visited[i])
dfs(i);
}
memset(visited, 0, sizeof(visited));
cnt = 0;
for(int i = n * m - 1; i >= 0; i--)
{
if(!visited[order[i]])
{
cnt++;
dfs_rev(order[i]);
}
}
e = 0;
for(int i = 0; i < num; i++)
{
int u = id[uu[i]];
int v = id[vv[i]];
if(u != v) newinsert(u, v);
}
ans = 0;
for(int u = cnt; u >= 1; u--)
{
int tmp = 0;
dp[u] = newval[u];
for(int i = newhead[u]; i != -1; i = newedge[i].next)
{
int v = newedge[i].v;
tmp = max(tmp, dp[v]);
}
dp[u] += tmp;
}
ans = dp[id[1]];
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &n, &m);
Kosaraju();
printf("%d\n", ans);
}
return 0;
}
另外附上tarjan的
偷懒用了vector
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#define MAXN 1666
#define MAXM 55555
#define INF 1000000000
using namespace std;
char mp[55][55];
int dfn[MAXN], low[MAXN], index, fa[MAXN], instack[MAXN];
int scc, n, m, cnt;
int s[MAXN], top, q[MAXM];
int val[MAXN], newval[MAXN], dp[MAXN], vis[MAXN];
int magic[MAXN];
struct Edge
{
int v, w;
Edge(){}
Edge(int a, int b){v = a; w = b;}
};
vector<int>g[MAXN];
vector<Edge>dag[MAXN];
void init()
{
index = top = scc = cnt = 0;
memset(dfn, 0, sizeof(dfn));
memset(instack, 0, sizeof(instack));
memset(newval, 0, sizeof(newval));
memset(val, 0, sizeof(val));
for(int i = 0; i < MAXN; i++)
g[i].clear(), dag[i].clear();
}
void tarjan(int u)
{
dfn[u] = low[u] = ++index;
s[++top] = u;
instack[u] = 1;
int size = g[u].size();
int v;
for(int i = 0; i < size; i++)
{
v = g[u][i];
if(!dfn[v])
{
tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(instack[v]) low[u] = min(low[u], dfn[v]);
}
if(dfn[u] == low[u])
{
scc++;
do
{
v = s[top--];
instack[v] = 0;
fa[v] = scc;
newval[scc] += val[v];
}while(v != u);
}
}
void build()
{
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
{
int u = i * m + j;
if(mp[i][j] == '#') {val[u] = -1; continue;}
else if(mp[i][j] == '*') magic[cnt++] = u;
else val[u] = mp[i][j] - '0';
if(j < m - 1 && mp[i][j + 1] != '#') g[u].push_back(u + 1);
if(i < n - 1 && mp[i + 1][j] != '#') g[u].push_back(u + m);
}
int x, y;
for(int i = 0; i < cnt; i++)
{
scanf("%d%d", &x, &y);
int u = x * m + y;
if(val[u] == -1) continue;
g[magic[i]].push_back(u);
}
}
void spfa(int src)
{
for(int i = 0; i <= scc; i++) dp[i] = -INF;
memset(vis, 0, sizeof(vis));
vis[src] = 1;
int h = 0, t = 0;
q[t++] = src;
dp[src] = 0;
while(h < t)
{
int u = q[h++];
vis[u] = 0;
int size = dag[u].size();
for(int i = 0; i < size; i++)
{
int v = dag[u][i].v;
int w = dag[u][i].w;
if(dp[v] < dp[u] + w)
{
dp[v] = dp[u] + w;
if(!vis[v])
{
vis[v] = 1;
q[t++] = v;
}
}
}
}
}
void solve()
{
for(int i = 0; i < n * m; i++)
if(!dfn[i]) tarjan(i);
for(int u = 0; u < n * m; u++)
{
int size = g[u].size();
for(int j = 0; j < size; j++)
{
int v = g[u][j];
if(fa[u] != fa[v]) dag[fa[u]].push_back(Edge(fa[v], newval[fa[v]]));
}
}
dag[0].push_back(Edge(fa[0], newval[fa[0]]));
spfa(0);
int ans = 0;
for(int i = 0; i <= scc; i++)
ans = max(ans, dp[i]);
printf("%d\n", ans);
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &n, &m);
init();
for(int i = 0; i < n; i++) scanf("%s", mp[i]);
build();
solve();
}
return 0;
}
拓扑排序版本 之前一直错是因为在拓扑排序开始时，只把0结点属于的强连通分量加入了队列， 其实不该这样。要把所有入度为0的结点加入，然后正常的拓扑排序，得到序列后，逆序进行DP就可以得到结果
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#define MAXN 1666
#define MAXM 55555
#define INF 1000000000
using namespace std;
char mp[55][55];
int dfn[MAXN], low[MAXN], index, fa[MAXN], instack[MAXN];
int scc, n, m, cnt;
int s[MAXN], top, q[MAXM];
int val[MAXN], newval[MAXN], dp[MAXN], vis[MAXN];
int magic[MAXN], in[MAXN];
vector<int>g[MAXN];
vector<int>dag[MAXN];
void init()
{
index = top = scc = cnt = 0;
memset(dfn, 0, sizeof(dfn));
memset(instack, 0, sizeof(instack));
memset(newval, 0, sizeof(newval));
memset(val, 0, sizeof(val));
memset(in, 0, sizeof(in));
for(int i = 0; i < MAXN; i++)
g[i].clear(), dag[i].clear();
}
void tarjan(int u)
{
dfn[u] = low[u] = ++index;
s[++top] = u;
instack[u] = 1;
int size = g[u].size();
int v;
for(int i = 0; i < size; i++)
{
v = g[u][i];
if(!dfn[v])
{
tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(instack[v]) low[u] = min(low[u], dfn[v]);
}
if(dfn[u] == low[u])
{
scc++;
do
{
v = s[top--];
instack[v] = 0;
fa[v] = scc;
newval[scc] += val[v];
}while(v != u);
}
}
void build()
{
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
{
int u = i * m + j;
if(mp[i][j] == '#') {val[u] = -1; continue;}
else if(mp[i][j] == '*') magic[cnt++] = u;
else val[u] = mp[i][j] - '0';
if(j < m - 1 && mp[i][j + 1] != '#') g[u].push_back(u + 1);
if(i < n - 1 && mp[i + 1][j] != '#') g[u].push_back(u + m);
}
int x, y;
for(int i = 0; i < cnt; i++)
{
scanf("%d%d", &x, &y);
int u = x * m + y;
if(val[u] == -1) continue;
g[magic[i]].push_back(u);
}
}
void topsort()
{
int h = 0, t = 0;
for(int i = 1; i <= scc; i++)
if(in[i] == 0) q[t++] = i;
while(h < t)
{
int u = q[h++];
int size = dag[u].size();
for(int i = 0; i < size; i++)
{
int v = dag[u][i];
in[v]--;
if(in[v] == 0) q[t++] = v;
}
}
memset(dp, 0, sizeof(dp));
for(int i = t - 1; i >= 0; i--)
{
int u = q[i];
int tmp = 0;
dp[u] = newval[u];
int size = dag[u].size();
for(int j = 0; j < size; j++)
{
int v = dag[u][j];
tmp = max(tmp, dp[v]);
}
dp[u] += tmp;
}
printf("%d\n", dp[fa[0]]);
}
void solve()
{
for(int i = 0; i < n * m; i++)
if(!dfn[i]) tarjan(i);
for(int u = 0; u < n * m; u++)
{
int size = g[u].size();
for(int j = 0; j < size; j++)
{
int v = g[u][j];
if(fa[u] != fa[v]) {dag[fa[u]].push_back(fa[v]); in[fa[v]]++;}
}
}
topsort();
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &n, &m);
init();
for(int i = 0; i < n; i++) scanf("%s", mp[i]);
build();
solve();
}
return 0;
}
分享到：
相关推荐
POJ2186-Popular Cows ...【Tarjan+极大强连通分量+缩点】 解题报告+AC代码 http://hi.csdn.net/!s/BGDH68 附：我所有的POJ解题报告链接 . http://blog.csdn.net/lyy289065406/article/details/6642573
北大POJ3733-Changing Digits【DFS+强剪枝】 解题报告+AC代码
北大POJ3373-Changing Digits【DFS+强剪枝】 解题报告+AC代码
北大POJ3009-Curling 2.0【DFS+Vector+回溯+剪枝】 解题报告+AC代码
北大POJ3026-Borg Maze【BFS+Prim】 解题报告+AC代码
北大POJ初级-所有题目AC代码+解题报告
1.把自己的污水排到河里V 2.把自己的污水送到右边> 3.把自己的污水送到左边
POJ3177-Redundant Paths 【Tarjan-边双连通分量-缩点】 解题报告+AC代码+测试数据 http://hi.csdn.net/!s/GPAY6Z 附：我所有的POJ解题报告链接 . http://blog.csdn.net/lyy289065406/article/details/6642573
POJ3352-Road Construction 【Tarjan-边双连通分量-缩点】 解题报告+AC代码 http://hi.csdn.net/!s/0T8UO5 附：我所有的POJ解题报告链接 . http://blog.csdn.net/lyy289065406/article/details/6642573
关于poj dp分类，我一直寻找dp的分类，终于找到了，也上传一下
北大POJ初级-简单搜索 解题报告+AC代码
北大POJ初级题-数据结构：解题报告+AC代码
北大POJ2187-Beauty Contest 解题报告+AC代码
博客链接 http://blog.csdn.net/CABI_ZGX/article/details/52701138
北大POJ3252-Round Numbers 解题报告+AC代码
北大POJ1416-Shredding Company 解题报告+AC代码
北大POJ1163-The Triangle 解题报告+AC代码
北大POJ1019-Number Sequence 解题报告+AC代码
北大POJ3126-Prime Path 解题报告+AC代码
北大POJ1472-Instant Complexity 解题报告+AC代码