hdu1874最短路

裸裸的最短路问题，将while(scanf("%d%d", &N, &M)!=EOF)粗心写为while(scanf("%d%d", &N, &M),我还奇怪怎么一直是超时，OMG.

首先用Dijstra算法，寻找两点间最短路径长度，算法复杂度是O(n^2).

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
using namespace std;
#define MAX_SIZE 202
#define MAX_NUMBER INT_MAX/2
int dis[MAX_SIZE];
bool visit[MAX_SIZE];
int G[MAX_SIZE][MAX_SIZE];
int N, M;
void Dijstra(int s,int t);
int main() {
	int i, j,k,w,s,t;
	while (scanf("%d%d", &N, &M)!=EOF) {
		for (i = 0; i < N; i++) {
			G[i][i] = 0;
			for (j = i + 1; j < N; j++)
				G[i][j] = G[j][i] = MAX_NUMBER;
		}
		for (k = 0; k < M; k++) {
			scanf("%d%d%d", &i, &j, &w);
			if(w<G[i][j])
			   G[i][j] = G[j][i] = w;
		}
		scanf("%d%d", &s, &t);
		Dijstra(s,t);
		if (dis[t] != MAX_NUMBER)
			printf("%d\n", dis[t]);
		else
			printf("-1\n");
	}
	return 0;
}
void Dijstra(int s,int t) {
	int i, j, k,pos,lmin;
	for (i = 0; i < N; i++) {
		visit[i] = 0;
		dis[i] = MAX_NUMBER;
	}
	j = s;
	dis[j] = 0;
	visit[j] = 1;
	for (i = 0; i < N; i++) {
		for (k = 0; k <N; k++) {
			if (!visit[k] && dis[k]>dis[j] + G[k][j])
				dis[k] = dis[j] + G[k][j];
		}
		pos =s; lmin = MAX_NUMBER;
		for (k =0; k <N; k++) {
			if (!visit[k] && lmin > dis[k]) {
				pos = k;
				lmin = dis[k];
			}
		}
		j = pos;
		if (j ==s||j==t)
			return;
		visit[j] = 1;
	}
}

　　接着使用Floyd算法也可以解决，时间复杂度是O(n^3),当然在这里是多此一举，不需要求出任意两点间最短距离。

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
using namespace std;
#define MAX_SIZE 202
#define MAX_NUMBER INT_MAX/2
int dis[MAX_SIZE][MAX_SIZE];
int G[MAX_SIZE][MAX_SIZE];
int N, M;
void Floyd();
int main() {
    int i, j, k, w, s, t;
    while (scanf("%d%d", &N, &M) != EOF) {
        for (i = 0; i < N; i++) {
            G[i][i] = 0;
            for (j = i + 1; j < N; j++)
                G[i][j] = G[j][i] = MAX_NUMBER;
        }
        for (k = 0; k < M; k++) {
            scanf("%d%d%d", &i, &j, &w);
            if (w<G[i][j])
                G[i][j] = G[j][i] = w;
        }
        scanf("%d%d", &s, &t);
        Floyd();
        if (dis[s][t] != MAX_NUMBER)
            printf("%d\n", dis[s][t]);
        else
            printf("-1\n");
    }
    return 0;
}
void Floyd() {
    int i, j, k;
    for (i = 0; i < N; i++)
        for (j = 0; j < N; j++)
            dis[i][j] = G[i][j];
    for (k = 0; k < N; k++)
        for (i = 0; i < N; i++)
            for (j = 0; j < N; j++)
                if (dis[i][j]>dis[i][k] + dis[k][j])
                    dis[i][j] = dis[i][k] + dis[k][j];
}