K - Leapin' Lizards - HDU 2732(最大流)
题意:在一个迷宫里面有一些蜥蜴,这个迷宫有一些柱子组成的,并且这些柱子都有一个耐久值,每当一只蜥蜴跳过耐久值就会减一,当耐久值为0的时候这个柱子就不能使用了,每个蜥蜴都有一个最大跳跃值d,现在想知道有多少蜥蜴不能离开迷宫(跳出迷宫就可以离开了。)
输入描述:输入矩阵的行M和跳跃值D,接着输入两个矩阵(列数自己求),第一个矩阵是每个柱子的耐久值,下面一个矩阵有'L'的表示这个柱子上有一只蜥蜴。
分析:题目明白还是比较容易的,矩阵的点数是20*20,也就400个点,对每个有耐久值的柱子进行拆点,然后连接两个柱子(判断是否在跳跃范围内),源点和有蜥蜴的柱子相连,汇点和所有能一下跳出来的柱子相连。注意输出的时候单词的复数单数。
下面是AC代码。
=================================================================================================================================
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
const int MAXN = 1005;
const int oo = 1e9+7;
struct point
{
int x, y, flow;
bool Lizards;
}p[MAXN];
struct Edge{int v, flow, next;}edge[MAXN*MAXN];
int Head[MAXN], cnt, Layer[MAXN];
void InIt()
{
cnt = 0;
memset(Head, -1, sizeof(Head));
}
bool canLine(point a, point b, int d)
{///判断能否从a点跳到b点
int len = (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);
if(d*d >= len)
return true;
else
return false;
}
void AddEdge(int u, int v, int flow)
{
edge[cnt].v = v;
edge[cnt].flow = flow;
edge[cnt].next = Head[u];
Head[u] = cnt++;
swap(u, v);
edge[cnt].v = v;
edge[cnt].flow = 0;
edge[cnt].next = Head[u];
Head[u] = cnt++;
}
bool BFS(int start, int End)
{
memset(Layer, 0, sizeof(Layer));
queue<int> Q; Q.push(start);
Layer[start] = 1;
while(Q.size())
{
int u = Q.front();Q.pop();
if(u == End)return true;
for(int j=Head[u]; j!=-1; j=edge[j].next)
{
int v = edge[j].v;
if(!Layer[v] && edge[j].flow)
{
Layer[v] = Layer[u] + 1;
Q.push(v);
}
}
}
return false;
}
int DFS(int u, int MaxFlow, int End)
{
if(u == End)return MaxFlow;
int uflow = 0;
for(int j=Head[u]; j!=-1; j=edge[j].next)
{
int v = edge[j].v;
if(Layer[v]-1==Layer[u] && edge[j].flow)
{
int flow = min(MaxFlow-uflow, edge[j].flow);
flow = DFS(v, flow, End);
edge[j].flow -= flow;
edge[j^1].flow += flow;
uflow += flow;
if(uflow == MaxFlow)
break;
}
}
if(uflow == 0)
Layer[u] = 0;
return uflow;
}
int Dinic(int start, int End)
{
int MaxFlow = 0;
while( BFS(start, End) == true )
MaxFlow += DFS(start, oo, End);
return MaxFlow;
}
int main()
{
int T, t=1;
scanf("%d", &T);
while(T--)
{
int i, j, N, M, d, k=0, sum=0;
char s[MAXN];
InIt();
scanf("%d%d", &M, &d);
for(i=0; i<M; i++)
{
scanf("%s", s);
N = strlen(s);
for(j=0; s[j]; j++)
{
k++;
p[k].x = i;
p[k].y = j;
p[k].flow = s[j]-'0';
}
}
for(i=k=0; i<M; i++)
{
scanf("%s", s);
for(j=0; s[j]; j++)
{
k++;
if(s[j] == 'L')
p[k].Lizards = true;
else
p[k].Lizards = false;
}
}
int start = k*2+1, End = start+1;
for(i=1; i<=k; i++)
{
if( p[i].flow > 0)
{
AddEdge(i, i+k, p[i].flow);
if(p[i].x-d<0 || p[i].x+d>=M || p[i].y-d<0 || p[i].y+d>=N)
{///能从这点跳出去,与汇点相连
AddEdge(i+k, End, oo);
}
if(p[i].Lizards)
{///如果这点是一个蜥蜴,与源点相连,流量是1
AddEdge(start, i, 1);
sum++;
}
for(j=1; j<=k; j++)
{
if(i == j)continue;
///如果能从i跳到j那么让他们相连,注意是拆点i与j相连
if(p[j].flow && canLine(p[i], p[j], d))
AddEdge(i+k, j, oo);
}
}
}
int Flow = Dinic(start, End);
Flow = sum - Flow;///要求逃不出来的
if(Flow == 0)
printf("Case #%d: no lizard was left behind.\n", t++);
else if(Flow == 1)
printf("Case #%d: 1 lizard was left behind.\n", t++);
else
printf("Case #%d: %d lizards were left behind.\n", t++, Flow);
}
return 0;
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
const int MAXN = 1005;
const int oo = 1e9+7;
struct point
{
int x, y, flow;
bool Lizards;
}p[MAXN];
struct Edge{int v, flow, next;}edge[MAXN*MAXN];
int Head[MAXN], cnt, Layer[MAXN];
void InIt()
{
cnt = 0;
memset(Head, -1, sizeof(Head));
}
bool canLine(point a, point b, int d)
{///判断能否从a点跳到b点
int len = (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);
if(d*d >= len)
return true;
else
return false;
}
void AddEdge(int u, int v, int flow)
{
edge[cnt].v = v;
edge[cnt].flow = flow;
edge[cnt].next = Head[u];
Head[u] = cnt++;
swap(u, v);
edge[cnt].v = v;
edge[cnt].flow = 0;
edge[cnt].next = Head[u];
Head[u] = cnt++;
}
bool BFS(int start, int End)
{
memset(Layer, 0, sizeof(Layer));
queue<int> Q; Q.push(start);
Layer[start] = 1;
while(Q.size())
{
int u = Q.front();Q.pop();
if(u == End)return true;
for(int j=Head[u]; j!=-1; j=edge[j].next)
{
int v = edge[j].v;
if(!Layer[v] && edge[j].flow)
{
Layer[v] = Layer[u] + 1;
Q.push(v);
}
}
}
return false;
}
int DFS(int u, int MaxFlow, int End)
{
if(u == End)return MaxFlow;
int uflow = 0;
for(int j=Head[u]; j!=-1; j=edge[j].next)
{
int v = edge[j].v;
if(Layer[v]-1==Layer[u] && edge[j].flow)
{
int flow = min(MaxFlow-uflow, edge[j].flow);
flow = DFS(v, flow, End);
edge[j].flow -= flow;
edge[j^1].flow += flow;
uflow += flow;
if(uflow == MaxFlow)
break;
}
}
if(uflow == 0)
Layer[u] = 0;
return uflow;
}
int Dinic(int start, int End)
{
int MaxFlow = 0;
while( BFS(start, End) == true )
MaxFlow += DFS(start, oo, End);
return MaxFlow;
}
int main()
{
int T, t=1;
scanf("%d", &T);
while(T--)
{
int i, j, N, M, d, k=0, sum=0;
char s[MAXN];
InIt();
scanf("%d%d", &M, &d);
for(i=0; i<M; i++)
{
scanf("%s", s);
N = strlen(s);
for(j=0; s[j]; j++)
{
k++;
p[k].x = i;
p[k].y = j;
p[k].flow = s[j]-'0';
}
}
for(i=k=0; i<M; i++)
{
scanf("%s", s);
for(j=0; s[j]; j++)
{
k++;
if(s[j] == 'L')
p[k].Lizards = true;
else
p[k].Lizards = false;
}
}
int start = k*2+1, End = start+1;
for(i=1; i<=k; i++)
{
if( p[i].flow > 0)
{
AddEdge(i, i+k, p[i].flow);
if(p[i].x-d<0 || p[i].x+d>=M || p[i].y-d<0 || p[i].y+d>=N)
{///能从这点跳出去,与汇点相连
AddEdge(i+k, End, oo);
}
if(p[i].Lizards)
{///如果这点是一个蜥蜴,与源点相连,流量是1
AddEdge(start, i, 1);
sum++;
}
for(j=1; j<=k; j++)
{
if(i == j)continue;
///如果能从i跳到j那么让他们相连,注意是拆点i与j相连
if(p[j].flow && canLine(p[i], p[j], d))
AddEdge(i+k, j, oo);
}
}
}
int Flow = Dinic(start, End);
Flow = sum - Flow;///要求逃不出来的
if(Flow == 0)
printf("Case #%d: no lizard was left behind.\n", t++);
else if(Flow == 1)
printf("Case #%d: 1 lizard was left behind.\n", t++);
else
printf("Case #%d: %d lizards were left behind.\n", t++, Flow);
}
return 0;
}
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