标签:des style blog http io ar color os sp
2263 题目意思是求起点城市到终点城市的途径中的最大载重量,可以用Dijkstra或者floyd 来解决,我是用的floyd 感觉更直观 因为只需要将递推式改成w[i][j] = Max(w[i][j],Min(w[i][k],w[k][j]));便可得到答案,而且floyd写法比较简单但是复杂度为n的3次方,不过此题还是能AC
Source Code Problem: 2263 User: lyz963254 Memory: 760K Time: 32MS Language: G++ Result: Accepted Source Code #include <iostream> #include <cstdio> #include <cstring> using namespace std; int Max(int x,int y) { return x>y?x:y; } int Min(int x,int y) { return x<y?x:y; } const int M = 250; int kase = 0; int n,r; int num; int w[M][M]; char city[M][35]; char start[35],dest[35]; int index(char *s) { int i; for(i=0;i<num;i++){ if(!strcmp(city[i],s)) return i; } strcpy(city[i],s); num++; return i; } int read() { int i,j,k,limit; scanf("%d%d",&n,&r); if(n==0) return 0; for(i=0;i<n;i++){ for(j=0;j<n;j++) w[i][j] = 0; w[i][i] = 9999999; } num = 0; for(k=0;k<r;k++){ scanf("%s%s%d",start,dest,&limit); i = index(start); j = index(dest); w[i][j] = w[j][i] = limit; } scanf("%s%s",start,dest); return 1; } void solve() { int i,j,k; for(k=0;k<n;k++){ for(i=0;i<n;i++){ for(j=0;j<n;j++){ w[i][j] = Max(w[i][j],Min(w[i][k],w[k][j])); } } } i = index(start); j = index(dest); printf("Scenario #%d\n%d tons\n\n",++kase,w[i][j]); } int main () { while(read()){ solve(); } return 0; }
1556 简直建图建醉了,算法直接套用bellman-ford 但是建图的过程太恶心了,需要判断墙的是否阻断了a,b顶点,阻断则不连边否则连边。。。
#include <cstdio> #include <iostream> #include <cstring> #include <cmath> using namespace std; #define inf 20000000000 #define M 150 struct Point { double x; double y; }p[M]; struct edge { int u; int v; }e[10050]; int n;//房间的数目 int pSize; //点的数目 double wX[20];//墙的X坐标 double pY[20][4];//Y的坐标 double map[M][M];//邻接矩阵 int eSize; //边的数目 int i,j; double Dis(Point a,Point b) { return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y)); } double Cross(double x1,double y1,double x2,double y2,double x3,double y3) //判断点是否在直线的上方还是下方 { return (x2-x1)*(y3-y1) - (x3-x1)*(y2-y1); } bool Ok(Point a,Point b) //判断在构造有向网的时候两点能否连边 { if(a.x>=b.x) return false; bool flag = true; int i = 0; while(wX[i]<=a.x&&i<n){ i++; } while(wX[i]<b.x&&i<n){ if(Cross(a.x,a.y,b.x,b.y,wX[i],0)*Cross(a.x,a.y,b.x,b.y,wX[i],pY[i][0])<0 || Cross(a.x,a.y,b.x,b.y,wX[i],pY[i][1])*Cross(a.x,a.y,b.x,b.y,wX[i],pY[i][2])<0 || Cross(a.x,a.y,b.x,b.y,wX[i],pY[i][3])*Cross(a.x,a.y,b.x,b.y,wX[i],10)<0){ flag = false; break; } i++; } return flag; } double bellman(int beg,int end) { double dist[M]; int i,j; for(i=0;i<M;i++){ dist[i] = inf; } dist[beg] = 0; bool ex = true; for(i=0;i<pSize&&ex;i++){ ex = false; for(j=0;j<eSize;j++){ if(dist[e[j].u]<inf&&(dist[e[j].v]>(dist[e[j].u]+map[e[j].u][e[j].v]))){ dist[e[j].v] = dist[e[j].u]+map[e[j].u][e[j].v]; ex = true; } } } return dist[end]; } void solve() { p[0].x = 0; p[0].y = 5; pSize = 1; for(i=0;i<n;i++){ scanf("%lf",&wX[i]); for(j=0;j<4;j++){ p[pSize].x = wX[i]; scanf("%lf",&p[pSize].y); pY[i][j] = p[pSize].y; pSize++; } } p[pSize].x=10; p[pSize].y=5; pSize++; for(i=0;i<pSize;i++){ for(j=0;j<pSize;j++){ map[i][j] = inf; } } eSize = 0; for(i=0;i<pSize;i++){ for(j=i+1;j<pSize;j++){ if(Ok(p[i],p[j])){ map[i][j] = Dis(p[i],p[j]); e[eSize].u = i; e[eSize].v = j; eSize++; } } } printf("%.2lf\n",bellman(0,pSize-1)); } int main () { while(scanf("%d",&n)!=EOF){ if(n==-1) break; solve(); } return 0; }
标签:des style blog http io ar color os sp
原文地址:http://www.cnblogs.com/lmlyzxiao/p/4136087.html