题意：题目是说一个n*m的迷宫中，有每个格子有柱子。柱子高度为0~3，高度为0的柱子是不能站的（高度为0就是没有柱子）在一些有柱子的格子上有一些蜥蜴，一次最多跳距离d,相邻格子的距离是1，只要跳出迷宫就是安全的。这个距离是曼哈顿距离（好像是的）。蜥蜴一次最多跳距离d,但是起跳的地方的柱子高度会减一，一个柱子同一时间只能有一个蜥蜴要求最少几个不能逃出迷宫。
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看懂了，明天拍
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 #include<queue> 7 #include<map> 8 using namespace std; 9 #define MOD 1000000007 10 #define pb(a) push_back(a) 11 const int INF=0x3f3f3f3f; 12 const double eps=1e-5; 13 typedef long long ll; 14 #define cl(a) memset(a,0,sizeof(a)) 15 #define ts printf("***** "); 16 const int MAXN=2200; 17 int n,m,tt,cnt; 18 char s1[30][30],s2[30][30]; 19 struct Edge 20 { 21 int to,next,cap,flow; 22 }edge[200020];//注意是MAXM 23 int tol; 24 int head[MAXN]; 25 int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN]; 26 int maze[30][30]; 27 void init() 28 { 29 tol = 0; 30 memset(head,-1,sizeof(head)); 31 } 32 //加边，单向图三个参数，双向图四个参数 33 void addedge(int u,int v,int w,int rw=0) 34 { 35 edge[tol].to = v;edge[tol].cap = w;edge[tol].next = head[u]; 36 edge[tol].flow = 0;head[u] = tol++; 37 edge[tol].to = u;edge[tol].cap = rw;edge[tol].next = head[v]; 38 edge[tol].flow = 0;head[v]=tol++; 39 } 40 //输入参数：起点、终点、点的总数 41 //点的编号没有影响，只要输入点的总数 42 int sap(int start,int end,int N) 43 { 44 memset(gap,0,sizeof(gap)); 45 memset(dep,0,sizeof(dep)); 46 memcpy(cur,head,sizeof(head)); 47 int u = start; 48 pre[u] = -1; 49 gap[0] = N; 50 int ans = 0; 51 while(dep[start] < N) 52 { 53 if(u == end) 54 { 55 int Min = INF; 56 for(int i = pre[u];i != -1; i = pre[edge[i^1].to]) 57 if(Min > edge[i].cap - edge[i].flow) 58 Min = edge[i].cap - edge[i].flow; 59 for(int i = pre[u];i != -1; i = pre[edge[i^1].to]) 60 { 61 edge[i].flow += Min; 62 edge[i^1].flow -= Min; 63 } 64 u = start; 65 ans += Min; 66 continue; 67 } 68 bool flag = false; 69 int v; 70 for(int i = cur[u]; i != -1;i = edge[i].next) 71 { 72 v = edge[i].to; 73 if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]) 74 { 75 flag = true; 76 cur[u] = pre[v] = i; 77 break; 78 } 79 } 80 if(flag) 81 { 82 u = v; 83 continue; 84 } 85 int Min = N; 86 for(int i = head[u]; i != -1;i = edge[i].next) 87 if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min) 88 { 89 Min = dep[edge[i].to]; 90 cur[u] = i; 91 } 92 gap[dep[u]]--; 93 if(!gap[dep[u]])return ans; 94 dep[u] = Min+1; 95 gap[dep[u]]++; 96 if(u != start) u = edge[pre[u]^1].to; 97 } 98 return ans; 99 } 100 int main() 101 { 102 int i,j,k; 103 #ifndef ONLINE_JUDGE 104 freopen("1.in","r",stdin); 105 #endif 106 scanf("%d",&tt); 107 int ca=0; 108 int d; 109 while(tt--) 110 { 111 ca++; 112 init(); 113 scanf("%d%d",&n,&d); 114 for(i=0;i<n;i++) 115 { 116 scanf("%s",&s1[i]); 117 } 118 m=strlen(s1[0]); 119 int tot=0; 120 cl(maze); 121 for(i=0;i<n;i++) 122 { 123 for(j=0;j<m;j++) 124 { 125 if(s1[i][j]>'0') //说明改点可跳 126 { 127 maze[i][j]=++tot; 128 addedge(2*tot-1,2*tot,s1[i][j]-'0'); 129 } 130 } 131 } //这里的tot要注意，如果一开始是tot++，那么比实际会多出一个点 132 int start=0,end=2*tot+1,nodenum=2*tot+2; 133 int sum=0; 134 for(i=0;i<n;i++) 135 { 136 scanf("%s",s2[i]); 137 for(j=0;j<m;j++) 138 { 139 if(s2[i][j]=='L') 140 { 141 sum++; 142 addedge(start,2*maze[i][j]-1,1); 143 } 144 } 145 } 146 for(i=0;i<n;i++) 147 { 148 for(j=0;j<m;j++) 149 { 150 if(maze[i][j]) //能直接跳出边界的点连上汇点 151 { 152 for(int x=-d;x<=d;x++) 153 { 154 for(int y=fabs(x)-d;y<=d-fabs(x);y++) //新技能get 155 { 156 int ni=i+x; 157 int nj=j+y; 158 if(ni<0||ni>=n||nj<0||nj>=m)continue; 159 if(ni==i&&nj==j) continue; 160 if(maze[ni][nj]==0) continue; 161 addedge(2*maze[i][j],2*maze[ni][nj]-1,INF); //这里的跳跃是无限流的 162 } 163 } 164 if(i<d||j<d||n-i<=d||m-j<=d) addedge(2*maze[i][j],end,INF); 165 } 166 } 167 } 168 int ans=sum-sap(start,end,nodenum); 169 if(ans==0)printf("Case #%d: no lizard was left behind. ",ca); 170 else if(ans==1)printf("Case #%d: 1 lizard was left behind. ",ca); 171 else printf("Case #%d: %d lizards were left behind. ",ca,ans); 172 } 173 }
1 //============================================================================ 2 // Name : HDU.cpp 3 // Author : 4 // Version : 5 // Copyright : Your copyright notice 6 // Description : Hello World in C++, Ansi-style 7 //============================================================================ 8 9 #include <iostream> 10 #include <string.h> 11 #include <stdio.h> 12 #include <algorithm> 13 using namespace std; 14 15 const int MAXN=2200; 16 const int MAXM=200020; 17 const int INF=0x3f3f3f3f; 18 struct Node 19 { 20 int to,next,cap; 21 }edge[MAXM];//注意是MAXM 22 int tol; 23 int head[MAXN]; 24 int gap[MAXN],dis[MAXN],pre[MAXN],cur[MAXN]; 25 26 void init() 27 { 28 tol=0; 29 memset(head,-1,sizeof(head)); 30 } 31 void addedge(int u,int v,int w,int rw=0) 32 { 33 edge[tol].to=v;edge[tol].cap=w;edge[tol].next=head[u];head[u]=tol++; 34 edge[tol].to=u;edge[tol].cap=rw;edge[tol].next=head[v];head[v]=tol++; 35 } 36 int sap(int start,int end,int nodenum) 37 { 38 memset(dis,0,sizeof(dis)); 39 memset(gap,0,sizeof(gap)); 40 memcpy(cur,head,sizeof(head)); 41 int u=pre[start]=start,maxflow=0,aug=-1; 42 gap[0]=nodenum; 43 while(dis[start]<nodenum) 44 { 45 loop: 46 for(int &i=cur[u];i!=-1;i=edge[i].next) 47 { 48 int v=edge[i].to; 49 if(edge[i].cap&&dis[u]==dis[v]+1) 50 { 51 if(aug==-1||aug>edge[i].cap) 52 aug=edge[i].cap; 53 pre[v]=u; 54 u=v; 55 if(v==end) 56 { 57 maxflow+=aug; 58 for(u=pre[u];v!=start;v=u,u=pre[u]) 59 { 60 edge[cur[u]].cap-=aug; 61 edge[cur[u]^1].cap+=aug; 62 } 63 aug=-1; 64 } 65 goto loop; 66 } 67 } 68 int mindis=nodenum; 69 for(int i=head[u];i!=-1;i=edge[i].next) 70 { 71 int v=edge[i].to; 72 if(edge[i].cap&&mindis>dis[v]) 73 { 74 cur[u]=i; 75 mindis=dis[v]; 76 } 77 } 78 if((--gap[dis[u]])==0)break; 79 gap[dis[u]=mindis+1]++; 80 u=pre[u]; 81 } 82 return maxflow; 83 } 84 85 char g1[30][30]; 86 char g2[30][30]; 87 int mat[30][30]; 88 89 int main() 90 { 91 //freopen("in.txt","r",stdin); 92 //freopen("out.txt","w",stdout); 93 int T; 94 int n,d; 95 int iCase=0; 96 scanf("%d",&T); 97 while(T--) 98 { 99 iCase++; 100 scanf("%d%d",&n,&d); 101 init(); 102 int tol=0; 103 for(int i=0;i<n;i++) 104 scanf("%s",&g1[i]); 105 int m=strlen(g1[0]); 106 memset(mat,0,sizeof(mat)); 107 for(int i=0;i<n;i++) 108 for(int j=0;j<m;j++) 109 if(g1[i][j]>'0') 110 { 111 mat[i][j]=++tol; 112 addedge(2*tol-1,2*tol,g1[i][j]-'0'); 113 } 114 115 int start=0,end=2*tol+1,nodenum=2*tol+2; 116 //进行拆点，加上源点和汇点，共2*tol+2个点。 117 int sum=0;//总数 118 for(int i=0;i<n;i++) 119 { 120 scanf("%s",&g2[i]); 121 for(int j=0;j<m;j++) 122 if(g2[i][j]=='L') 123 { 124 sum++; 125 addedge(start,2*mat[i][j]-1,1); 126 } 127 } 128 for(int i=0;i<n;i++) 129 for(int j=0;j<m;j++) 130 if(mat[i][j]) 131 { 132 for(int x=-d;x<=d;x++) 133 for(int y=abs(x)-d;y<=d-abs(x);y++) 134 { 135 int newi=i+x; 136 int newj=j+y; 137 if(newi<0||newi>=n||newj<0||newj>=m)continue; 138 if(mat[newi][newj]==0)continue; 139 if(newi==i&&newj==j)continue; 140 addedge(2*mat[i][j],2*mat[newi][newj]-1,INF); 141 } 142 if(i<d||j<d||n-i<=d||m-j<=d) 143 addedge(2*mat[i][j],end,INF); 144 } 145 int ans=sum-sap(start,end,nodenum); 146 if(ans==0)printf("Case #%d: no lizard was left behind. ",iCase); 147 else if(ans==1)printf("Case #%d: 1 lizard was left behind. ",iCase); 148 else printf("Case #%d: %d lizards were left behind. ",iCase,ans); 149 } 150 return 0; 151 }